geometry-and-trigonometry

[signal field]

ok one sec im getting the screenshot

this is central arcs and angles my question is when would i use 2pir

i understand the formula theta/360=S/C

but i dont understand when would i use s/2pir=theta/360

this textbook im reading from doesnt really explain it well

[kind rapids]

Do you know C is the circumference of the circle?

[signal field]

yea

circumference/diameter

=2pr?

[kind rapids]

No, $C = 2 \pi r$

[somber coyote]

Closer

[signal field]

ok circumference =2pir

@kind rapids if i was to just say s/2pir for the first example would that be wrong?

instead of s/c

since c= 2pir

[kind rapids]

Well, no

But you are trying to compute circumference

So no need to make it to 2pi r

But if you need to compute r then you need to write it as 2pi r

[signal field]

ok so when i need to find radius use 2pir

[kind rapids]

For example in the first image you can make it to: $2\pi r = 24000$

[somber coyote]

Closer

[kind rapids]

And then find r aka radius

[signal field]

ok thank you!

ima have to get a on-campus tutor asap because its only been a week and im alr lost

[void ivy]

why is the answer 20? i was thinking 10

this is the answer but isn't that the just the whole area of the flag

[north kindle]

are we assuming the flag is a rectangle?

[void ivy]

sorry, triangle

the topic is about triangles

but now that you've mentioned it, it may be a rectangular flag, where the diagonal divides it into 2 triangles, which one is of area 20in^2

thanks 👍

[faint pasture]

And yea it cant be a triangle as a triangle doesnt have diagonals

[void ivy]

good point, i assumed it meant the 90º angle pointing at the hypotenuse, but it seems i was mistaken, thanks!

[faint pasture]

Well thatd be an altitude

[void ivy]

oh does it? hoepfully i get to that part soon, on the book i'm reading. i'm relearning math from the absolute basics to build a solid foundation before i move to more advanced topics like chemistry, physics, pre-calculus, and hopefully i'll understand calculus intuitively one day 🤣

[faint pasture]

Kool man

Calc isnt as tough as an high school tchr/ the society makes it out to be anyways

Its rather doable

Good luck though

[maiden brook]

what’s angle EAD

[mellow quest]

$\angle BAD= \angle BAE + \angle EAD \
90 = \angle BAE + 60$

[somber coyote]

olivka

[maiden brook]

what’s angle BEC then

[mellow quest]

150

[onyx tiger]

The higher the angle, the larger the opposite side

[tepid axle]

what does it mean when a function is defined for all R and how do i explain it?

i have two functions, f(x) and g(x) where h(x) = f(g(x)). f(x) and g(x) are both defined for all R

[faint pasture]

Then h(x) is defined for all R

[tepid axle]

yes my teacher said i didnt explain it right

and he wants me to explain more lol

he gave me a tip and wrote "is h(x) definition dependent on definition for f" - sry direct translation

[upper karma]

wdym

[tepid axle]

so he asked me if h domain is dependent on f domain since h(x) = f(g(x)). As both f and g are defined for all R. cause i just wrote h is defined for all R

$h(x)=-\frac{3\sin(\pi*\frac{3x}{2})}{4}-1$

[somber coyote]

delila

[pearl bolt]

Prove Sin^2 + cos^2 = 1 please help me out..

[obsidian harness]

from what definitions?

[pearl bolt]

What defination?

[obsidian harness]

normally we define cos to the the x-coordinate of the unit circle and sin to be the y-coordinate

so applying the Pythagorean theorem directly, cos^2 + sin^2 = 1

1^2 = 1

[pearl bolt]

But image is wrong

[obsidian harness]

howso

[pearl bolt]

cos should me in sin's place and sin should be in cos's place

@obsidian harness

And can you explain the image please

[obsidian harness]

I just explained it

[pearl bolt]

Oh yes

But how can we use pythagoreas here

I am 10th not something like university level

[obsidian harness]

cause we have a right angle between cos and sin

cos is the x-axis

sin is parallel to the y-axis

x- and y-axes are perpendicular

[pearl bolt]

And you get hypotenus which is 1

But how can it be 1

[obsidian harness]

yeah we define it to be the unit circle, so we start out with the circle x^2 + y^2 = 1^2

and then draw cos and sin on the unit circle

[pearl bolt]

Thank you very much mann

[faint pasture]

Another way is start out with pythogorean theorem

P²+ b² = h²

And divide by h²

Tho this only works for right triangles

But eh in 10th grade thats prolly all youre gonna have to deal with

[obsidian harness]

npnp

[obsidian harness]

also like you know, (opp/hyp)^2 + (adj/hyp)^2 = (opp^2 + adj^2)/hyp^2 = hyp^2 / hyp^2 = 1

by Pythagoras

I'm just introducing the unit circle cause things become simpler if we just let the hypotenuse be 1

and it's useful to do your trig in all 4 quadrants

rather than having to memorise CAST

[faint pasture]

But like idk if they were introduced to trig the unit circle way in schl

Cuz for me it was all triangles

So yea

I mean yes unit circle is objectively better

[obsidian harness]

I feel they should teach both at the same time

[faint pasture]

But yea

[obsidian harness]

like say for any triangle we can use SOHCAHTOA, but by similarity we can let hyp = 1

[faint pasture]

I mean the circle covers the triangle part anyways, but the vice versa is not true

[obsidian harness]

cause trig ratios aren't affected in similar shapes

[faint pasture]

So circle should be prioritised

But yea the ed system sucks here

[kind rapids]

I thought they teach trig in terms of triangles first and then to the unit circle

[faint pasture]

Yea thats what happened for me at least

[upper karma]

they do

[faint pasture]

But the unit circle came like an year later after the triangle shit

So yea

[upper karma]

but unit circle is just good

[faint pasture]

Yea its better objectively

[upper karma]

i was happy to visualize trig with unit circle

when i was taught trig

than i was with triangles

[tepid axle]

whats the difference between range and codomain? i wrote two different things on them but still got wrong 😦

[kind rapids]

I think they are the same?

[tepid axle]

oh my teacher replied and said that they in general dont have to be the same? 🫣

[obsidian harness]

the range, which you should refer to as the image, is a subset of the codomain

for example, if I have $y = x^2$ where $x, y \in \mathbb R$

[somber coyote]

south

[obsidian harness]

the codomain is R

[tepid axle]

yeah

[obsidian harness]

but the range is just f(x) >= 0

fixed the typo sorry

[tepid axle]

question

so is

[-1, 1] range then?

he said i had written basically the same on both

[obsidian harness]

for which function?

[tepid axle]

sin

just example xd

[obsidian harness]

I see this

oh

[obsidian harness]

yeah

[tepid axle]

yeah the range for the above one is smtn else

ohh ok idk he just wrote i had done the same on both

[obsidian harness]

yeah the codomain can be defined as anything, although a natural choice for the codomain is all real numbers

[tepid axle]

cause i had defined the function for all R and then for range i wrote tht

yeah

this is the range for the function above

[obsidian harness]

to properly define a function you need 3 things:

1. the domain
2. the codomain
3. the relation between the domain and codomain, the function 'rule'

at high school level we assume that 1) and 2) are all real numbers unless specified otherwise

[tepid axle]

ye

[obsidian harness]

not many high schools teach Diophantine equations

those are the solutions to ax + by = c if x, y are now integers

[tepid axle]

ohh i see

[obsidian harness]

so yeah 1) and 2) are important, just make a mental note of that

[tepid axle]

thanks so much

[obsidian harness]

np

[tepid axle]

also would u like to help me define function h heh D:

if u have time otherwise its ok

[obsidian harness]

I can't, thanks for asking tho

[pearl bolt]

Whats the point of taking sin and cos as y axis and x axis respectively and solving through Pythagoreous If we already knew that its hypo is 1.

Why should we consider hypo as its answer as 1?

Please help out with this

<@&286206848099549185>

[trail tendon]

you already know the hypotenuse is 1, and then using the unit circle you can see that cos(theta) and sin(theta) is defined to be the sides of the right triange. Thus, for any angle, sin^2(theta) + cos^2(theta) = the hypotenuse, and since you know the hypotenuse is 1 since it is a unit circle you know sin^2(theta) + cos^2(theta) = 1

[tepid axle]

to prove a function is surjective do i do f(x) = y?

and then calculate it?

[midnight merlin]

does anyone know the actual name of a cubic Torus(ring shape)??? im going crazy as it does not have a name

[azure helm]

help

[thorn dune]

Maybe 30?

[azure helm]

explain

AED and abcd are coplanar

[lyric sonnet]

Yes

[lyric sonnet]

Lie on the same flat surface

[azure helm]

yeah

it's 2d

[maiden brook]

we already answered this

[azure helm]

when

[azure helm]

60

[azure helm]

ye exactly

I figured it out

[azure helm]

help

[north kindle]

whats another way to express DAE

[azure helm]

I need more help

[atomic copper]

do i need to memorize trigonometric identities or can i just figure them out during my test

[light hemlock]

Hello, could anyone explain to me how I’m supposed to solve number 15. I already tried a few things but I’m getting the wrong answer still.

[warm shuttle]

you have b/c = sin(β)

[obsidian harness]

Using trig it's 2 * arctan(1/(sqrt3 + 2))

[azure helm]

how so?

can u make a drowing

[obsidian harness]

If M is the midpoint of DE, we want DM and AM

Let DM = 1 so the height of the square is 2

Then the height of the triangle will be sqrt3

So that arctan is angle DAM and we multiply it by 2

[oblique widget]

Can somebody help

[fallow palm]

It depends on how quickly you can derive them and how much time you have

[warm shuttle]

i don't like this

,w 2d distance formula

$\operatorname{dist}(a, b) = \sqrt{(x_a - x_b)^2 + (y_a - y_b)^2}$

[somber coyote]

CyclicTree

[warm shuttle]

this 4 times (or 2 if you are lazy) gives perimeter

,w gauss shoelace formula

$2A = (x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2) + \ldots + (x_n y_1 - y_n x_1)$

[somber coyote]

CyclicTree

[warm shuttle]

gives area

or you can find the height. or you can also use law of cosines or cross product or determinant or whatever else

[obsidian harness]

mixed up opposite and adjacent woops

[azure helm]

how do I get there aswell

[obsidian harness]

there has to be a simpler solution than using trig

[azure helm]

okay

geo is tricky

[obsidian harness]

oh wait I got it

CD = CB = CA cause ABC is equilateral, so triangle CDA is isosceles

so angle CAD = (180 - 90 - 60)/2 = 15

by symmetry, angle EAB = 15

so angle CAD + angle DAE + angle EAB = 60 and hence 15 + angle DAE + 15 = 60

DAE = 30

ok

[azure helm]

can you show in a drawing

i doont really get it

[obsidian harness]

why don't you try drawing it yourself

[azure helm]

i am having a bit of trouble with that

[obsidian harness]

ok, with which part exactly?

[azure helm]

everything I dont get it

you went too fast

[obsidian harness]

now you're blaming me?

[azure helm]

if you can go slower that would be fine

no sorry im am not

I just wanted you to help me a bit with the drawing thats it

is that fine for you, then I would nmot disturb you any longer

[obsidian harness]

is this better then

[azure helm]

how did u

do that

[obsidian harness]

I guess what I wrote isn't sinking in for you

and I don't think me speaking more will help

[oblique widget]

I literally cant find it

I found the perimeter but not the area

I used the distance formula

[warm shuttle]

Here's an idea

You see how the x axis splits it into two triangles?

Since parallelograms have opposite sides equal, you can cut off top triangle and move its long side to match the long side of the bottom triangle

but then you still have the slanted bit on the right

which you can again cut off and move to the left

now you have a 5x17 rectangle

so the area is 5*17

[upper karma]

https://en.wikipedia.org/wiki/Inscribed_figure

According to conventions then is it alright to call a circle “inscribed” in a rectangle?

cuz it doesn’t touch all its edge and can at most touch 3

[faint pasture]

It can touch all of them though?

[upper karma]

i mean yes obviously

[faint pasture]

A square is just a special kind of rectangle

[upper karma]

either 2 or at most 3

[faint pasture]

So i dont get what you mean

[upper karma]

if it’s a square then 4

[faint pasture]

It can touch all 4

[upper karma]

i’m asking about the definition of inscribed

and whether it allows shapes like this which isn’t “tangent” to all sides

[faint pasture]

Yea if it doesnt touch all the sides then its not inscribed simple as that

[upper karma]

where in the wiki does it say that

because i think this has to do with definitions

and conventions

[faint pasture]

To say that "figure F is inscribed in figure G" means precisely the same thing as "figure G is circumscribed about figure F"

Iguess so

But idk how this will extend to circles

[upper karma]

Yeah but if u search “circle inscribed in rectangle” u see our case

i need a definitive answer on what the convention is

[faint pasture]

I mean this kinda means that?

But what even is a vertex in the case of a circle

Tho there definitely doesnt lie one on the 4th side

So yea

[upper karma]

that’s different

that’s talking about inscribing a polygon in a circle

[faint pasture]

How so

[upper karma]

check the sentence before that

[faint pasture]

It clearly says the outer figure is a polygon so?

[upper karma]

no

A circle or ellipse inscribed in a convex polygon (or a sphere or ellipsoid inscribed in a convex polyhedron) is tangent to every side or face of the outer figure (but see Inscribed sphere for semantic variants).

[faint pasture]

They are not really related according to me

[upper karma]

this sentence answers it

you’re looking at the next

which is the “opposite”

as in, it’s circumscribed now not inscribed

[faint pasture]

Nvm why th did i miss that sorry for wasting your time

[faint pasture]

30?

Basically join AD and AE

And ACD and ABE (the two new triangles) are isoceless

Soo BAE = BEA

And we also know ABE =ABC+CBE =90+60 =150

So now you can find BAE and CAD

Also CAB = 60 = CAD +DAE + BAE

[outer wadi]

what does the second point mean

[faint pasture]

DI = FI = EI

[outer wadi]

oh so basically circle inside triangle touches point F,E,D

?

[upper karma]

the circle is tangent to each side yeah

[outer wadi]

thanks

[upper karma]

ender already provided u a solution, but if u wanted another one:

if you can show <CAN = <BAK = 15 then you're done

[azure helm]

how

[warm shuttle]

how do you know these are equal

[azure helm]

I have the same exact question

[warm shuttle]

this one is good tho

[azure helm]

help

[pearl bolt]

Prove sec

[warm shuttle]

DCB = 90°, BCA = 60° -> DCA = 150°.
CB = DC, CB = CA -> CD = CA -> CDA = CAD.
DCA + CDA + CAD = 180°, CDA = CAD -> DCA + 2CAD = 180°.
DCA = 150°, DCA + 2CAD = 180° -> 150° + 2CAD = 180° -> 2CAD = 30° -> CAD = 15°.
Symmetrically, BAE = 15°.
ABC - equilateral -> CAB = 60°.
CAD + DAE + EAB = CAB, CAB = 60°, CAD = BAE = 15° -> 15° + DAE + 15° = 60° -> DAE = 30°

[pearl bolt]

Prove sec^2 = 1+ tan^2

[warm shuttle]

what's sec?

sec(x) = 1/cos(x)
sec²(x) = 1/cos²(x)

tan(x) = sin(x)/cos(x)
1+tan²(x) = 1 + sin²(x)/cos²(x) = cos²(x)/cos²(x) + sin²(x)/cos²(x) = (cos²(x) + sin²(x))/cos²(x) = 1/cos²(x)

[upper karma]

they’re similar triangles

also C(red) = B(red)

[wraith hill]

can somebody please help me with trigonometry?

[obsidian harness]

problem 1: recall that a 45-45-90 triangle has sides 1 : 1 : sqrt(2)

so what must you have to get ? : ? : x, if the hypotenuse is x

once you figure out what ? is, ? + x + ? = 2 so you have an equation in x you can solve

[warm shuttle]

for problem 2 you can consider triangle formed by centers of 3 smaller circles

and then find the distance from its center to one of the vertices

that +a gives you radius of the bigger circle

[obsidian harness]

it's the same deal with problem 2

the middle triangle is equilateral cause all the sides have length 2a

so you can use trig to figure out what ?, the distance to the centre of the big circle and the centre of the equilateral triangle, must be

sine rule for example

or right angled trig

[wraith hill]

I appreciate the help but can you please make it clearer?

[obsidian harness]

what could I make clearer?

[wraith hill]

I didn't really get the terms you used to represent it

[wraith hill]

ok thanks that made it clearer

can I also ask how to get the sides of the triangle with only angles?

[obsidian harness]

it's a 45-90-45 triangle so you could use trig if you wanted to think about it that way

so x cos 45 or x sin 45 would both work

[wraith hill]

ok thanks

[tepid axle]

if my function is supposedly surjective, the codomain and range is the same?

[spiral widget]

Yes

[tepid axle]

$[-\frac{7}{4}, -\frac{1}{4}]$ so this is codomain and range?

[somber coyote]

delila

[spiral widget]

For what function

[tepid axle]

$h(x)=-\frac{3\sin(\pi*\frac{3x}{2})}{4}-1$

[somber coyote]

delila

[tepid axle]

where h(x) = f(g(x))

f is defined for all R and g R to R

[spiral widget]

So the range would be this

[tepid axle]

ya

[spiral widget]

The codomain would normally be part of the data of the function

In this case it could be any set containing the range

[tepid axle]

oh how do i

[spiral widget]

So, the codomain would normal be given to you, and it’s not really something you can deduce from just a formula

The range/image is, though, so long as you know the domain

[tepid axle]

yeah

my teacher said i got the codomain right but explanation wrong

for why domain is R and codomain

[lilac nebula]

can anyone help me to find that angle??

[faint hollow]

DAO is coming out to be 80 ig and DEC is 60

[lilac nebula]

thanks bro

[stiff crane]

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEggvi05DI_YHZxLd7sssoDiJWm0fEIW-lzKQM_xd74qC9EuAoXO3aBc0c__D9QWBfh9_FpEHgEolna64rZjmKj9VFqvClasJ1tc1lRZPF3aSD-qVJEH2VQOTlsiGKOsiqXrRxm9t5vHSg/s1600/7a+cristal+noir+-+un+horizon+et+quasi-cristaux.png

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhToK7iPOLy-NU1x5SbNzQqwka_ZicnmMrYr9H8tu__1XqbwGXFdSDHt7C2335FGNz5kQumCtoXEvxNJ96SKJkoKSiERqdPhe7Yr8EVni3Odt3vBFP0zg1DA4NW2QiitTgrxLrxGxE_Mg/s1600/Diffraction_Quasi_Cristaux.jpg

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg_lpO2bcZ9CRxbuZp3qNcYDbNr3tkCDCCpoXt7waVqYhnUdweWphxRRU4gBjSJNh9TtFN5uVLRxZS1siShmTvjgRucpZgX3khS7cx5B7F4wpL50sRC6CX2_6wgBBf0etaGIcQOyTwk5A/s1600/cristal+noir+11.png

Quasi-cristaux 3D

Good day!

[pearl bolt]

Solve x^2 + x + 1

[trail tendon]

theres nothing to solve ._.

[pearl bolt]

There is find roots

[warm shuttle]

,w x^2 + x + 1 find roots

[somber coyote]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by cyclic_tree

[faint pasture]

quadratic formula or the factorison of x^3 = 1 and thus x^3-1 = 0 gives that

[north kindle]

im having trouble with this problem, i cant find any remotely nice expression for this line

i got y = -x(2ca/(-2b + a))/a + 2ca/(-2b + a) but i feel like theres a better one

[faint pasture]

You could multiply both sides by (-2b + a) but that is the best we can do i guess

[oblique widget]

The top one

[spiral lodge]

Can you send a full picture?

Btw, is that a test you're doing right now?

[trail tendon]

it seems that you already did this, but you can use the fact that any two sides added up should be greater than the third side

[spiral lodge]

I thought it was a question on Pythagora's theorem

Even though the 90° angle isn' marked with the usual little square 🤔

[oblique widget]

No….

[oblique widget]

I’m aware

[trail tendon]

so then...whats the problem?

where are you stuck 💀

[oblique widget]

What’s the next step

I have to write it as an inequality

[signal field]

How could I find an acute angle theta if it’s arcsin or arccos

Problem 33

Nvm I got it

[steady vault]

lol

[blazing veldt]

If $\cot \theta =- \frac{7}{2}$ and $\cos \theta > 0$, is $\tan \theta= -\frac{2}{7}$?

[somber coyote]

Primordial

[twin spindle]

Hello, im doing hw on Trig equations, and im confused on how to get multiple solutions, I was able to figure out 1 solution to this problem, but not for the other ones

[upper karma]

would point H be considered coplanar and why

[faint hollow]

Yeah ig, why will it be not??

[faint hollow]

I am getting total 4 distinct solutions of t --> cos t = 0 gives t = π/2,3π/2 and cos t = 3/4 gives t = 40,320(approx), are these the solutions?

[warm shuttle]

coplanar to what?

If the question is as you stated, then the answer is "no, because you can't talk about an individual point being coplanar"

[opal sierra]

Can someone teach me trig

In dm

[scenic field]

Hi

Pls i need help

Dm ne

Me*

[dawn scaffold]

i too require further assistance in this subject

nvm i got it guys :DD

[faint pasture]

Thats #calculus

(Find slope of the tangents then use that to find to slope of the normal and we also know the point)

[faint comet]

Guys, for question B does the domain after being manipulated become -10pi/3,x,8pi/3. I multiplied by 3 first then minused pi/3

[graceful talon]

Nani

[zealous rapids]

like yes the domain for "3x-pi/3" sure

you manipulate the interval domain

-3pi <= 3x <= 3pi

so -10pi/3 <= 3x-pi/3 <= 8pi/3

[faint comet]

Can you find where u went wrong here

[faint hollow]

its hard to check this, but you can do one thing, just apply the general solution of cos θ which is --> θ = 2nπ ± α, where α ∈ (0, π], here is alpha is any solution in the given range(for example in your case alpha would be π/3) and n is set of integers, from this you can easily put values of n that gives you the values in [-10π/3, 8π/3]

then solve for x

[dawn carbon]

tips how to study basic trig fast im in alg2 and have a math comp with it soon

[twin spindle]

When converting from rectangular to polar, how do you know whether you should add 2pi, add pi, or not add anything at all? Because on some of these answers it tells you to only do one of them or either

[obsidian harness]

you should plot the point on the xy-plane to figure it out

so (-2, 1) is in the 2nd quadrant, so it would be pi - arctan(1/2) which is equivalent

the rule for arctan is that it always gives you an angle in the 1st or 4th quadrant

1st quadrant gives you a positive angle and 4th quadrant gives you a negative angle

so for the 3rd quadrant, add pi radians

for the 2nd quadrant, also add pi radians

for the 4th quadrant, you want the same position but you want to make it positive, so you add 2pi

[obsidian harness]

but instead of thinking about this as a list of rules, just plot the point and remember this

[twin spindle]

Alrr so basically we'd add pi or 2 pi, to find the other solution? And i'm guessing that the 2nd question didn't need to be added. because it's already in a positive coordinate?

[obsidian harness]

correct, (2, 4) is in the 1st quadrant, so we don't need to do anything

[twin spindle]

I see I see

What's wrong with having negative values tho?

I get if you were measuring something you don't want a negative value, cause that jus makes no sense.

But in a case like this, why change it to a positive, if you got sum from the 4th quadrant?

[obsidian harness]

there's nothing wrong, it's just that the question specifies $0 \le \theta < 2 \pi$

[somber coyote]

south

[obsidian harness]

that's the domain we usually use for angles

[twin spindle]

OHHHHHH

Omfg I can't believe I missed that 😭

That shi always gets me, whenever their's an interval

[obsidian harness]

*there's

[twin spindle]

Thank you for helping tho 🙏

[obsidian harness]

nw!

[pearl bolt]

Prove that sec^2 = 1+ tan^2 through circle method

Please help me to prove this

[next dome]

@everyone@here

1. Guys can anyone pls tell me that what is an arc in geometry?
2. Guys can anyone pls tell me that what is pi maths?

[modest mesa]

i need help with this

[stable ridge]

3sinA+5cosB=3
3cosA+5sinB=7
Find cot(a+b)
Pls help
I got Final exam tomorrow

[obsidian harness]

sin^2 + cos^2 = 1 right

now divide everything by cos^2

[obsidian harness]

don't ping everyone and here, they don't work

also you can Google the answers to your 2 questions

[obsidian harness]

find $\tan(a + b)$ first, so you can actually find $\sin(a + b)$ first

[somber coyote]

south

[obsidian harness]

square both of your simultaneous equations and add them together

you get $9 + 25 + 30 \sin a \cos b + 30 \cos a \sin b = 3^2 + 7^2 = 58$

[somber coyote]

south

[obsidian harness]

so use the Pythagorean identity to find cos(a + b)

great, now you have tan(a + b), then take the reciprocal of what you have to get cot(a + b)

[obsidian harness]

have you tried using the double angle identities for sin 2x on the LHS?

you should know how to express sin theta and cos theta in terms of tan theta at this stage

[modest mesa]

yeah on LHS i got some long term

but ig i did it?

[twin spindle]

Is it okay if someone checks my answers on this review paper?

The problem is that, I jus wrote my answers and did the work on paper

[agile cave]

Help

[hot latch]

can someone pls explain this work

please

[tall coral]

anywhere in particular you find yourself lost?

[hot latch]

how did she get the y value

and im just not seeing all the steps

[tall coral]

ok

the first step is listing the equivelences based on what tan is

then, they solve both equations for y

[hot latch]

so she just left off the "y="?

[tall coral]

yes

because if they're both y, they equal eachother

think you got it?

[hot latch]

so she set the both equal to eachother and simplified

[tall coral]

yep!

then plugged the x value into the original equation and solved for y

[hot latch]

where did the x in the right half of the equation go

oh

thats why its minus

no

idk

[tall coral]

$x \tan(63)- x \tan (53) \rightarrow x(\tan(63) - \tan (53))$

[hot latch]

OH

[somber coyote]

KirbysGames

[hot latch]

and then she divided

[tall coral]

yep!

[hot latch]

okay wow

and so i get the x value so how did she get 819

[tall coral]

chuck it back into the original equation

$\tan (63) = \frac{y}{417.5949}

[hot latch]

tan63=y/417 ?

[tall coral]

yep

[hot latch]

omg

[tall coral]

then solve y

[hot latch]

wow im slow

thank you

[tall coral]

you're good, we all have those moments lol.
and you're welcome

[hot latch]

have a great night

[quartz minnow]

I got a doubt in circles

[obsidian harness]

ask

[quartz minnow]

Why's there 2/3 ad= 2/3 BC and so on

[obsidian harness]

the centroid is always 2/3 of the way between a vertex and the midpoint of the opposite side

EM:DC = 1:2
so by similar triangles, EG:GC = 1:2

similarly for the other sides

cause it doesn't matter which points you label as A, B, C

[quartz minnow]

Thankz

[fringe perch]

450

[obsidian harness]

if you don't show your work this is what happens

you get ignored or someone just gives you the answer

without the solution

and don't expect people to give you a solution cause you could have pasted this question here in 2 secs and just left

without doing anything else

[fringe perch]

i had solved this question in the morning

[faint pasture]

Everyones willing to put just as much effort as you at max

[fringe perch]

anyone

i have a doubt

can anyone help me out

[obsidian harness]

just ask your q

[fringe perch]

i think the answer is b

@obsidian harness

[obsidian harness]

idk man

[fringe perch]

ok

why no one is responding btw?

[obsidian harness]

no one wants to do JEE stuff if it's not their homework

[faint hollow]

I actually like doing JEE stuff but I dislike pure geometry and 3d stuff,conics pretty good

[fringe perch]

Everyone does

[faint hollow]

Yeah if,it's pretty common, whoever I ask has the same opinion except some exceptions here and there

Btw which grade you in @fringe perch

[fringe perch]

11

You?

[faint hollow]

12

[fringe perch]

Noicee

[fringe perch]

But it's the one of the easiest chapters which fetches you marks

[faint hollow]

Yeah ik

[fringe perch]

Hm

[oblique widget]

Somebody please explain to me asap

[upper karma]

Perimeter is all the sides added up

[oblique widget]

SO I SET ALL THE EQUATION TO 20?

[upper karma]

NO

[upper karma]

NOT 20

22

ONE SEC

[oblique widget]

OH YEA

PMG

[upper karma]

ARE YOU TAKING A TEST?

[oblique widget]

NO BRUH

[upper karma]

Oh it's just practice

[oblique widget]

I wouldn’t be able to use my
Phone

[upper karma]

Yeah you're right

Sorry I'm kinda dumb

I don't actually know but I'll try

[oblique widget]

So can you help me

[upper karma]

Gimme a few mins

[oblique widget]

Ok

Did u get it

[upper karma]

Nope 😂

I'm trying

[fringe perch]

Equate ad with bc , and ab with dc

[oblique widget]

Ok

So

[upper karma]

It still gets messy

[oblique widget]

3x-2=x-w+1

Then

2y+1=3-4w

Right

[fringe perch]

Yes it would

[upper karma]

I got 71/9

[oblique widget]

Photo math said 7

[upper karma]

Man

What the flip

[fringe perch]

X is 2 and y is 3 it's seems

Ab is 5?

@oblique widget

[oblique widget]

IDK

[upper karma]

It's 7

[oblique widget]

HOW

HOW

[upper karma]

Trust photomath fr

[fringe perch]

Sry 7

[oblique widget]

HOW DO I FIND IT

[upper karma]

One sec

[fringe perch]

The source is correct

Il send u wait

[oblique widget]

OKK

[upper karma]

Wait photmath is wrong

[oblique widget]

Are you guys searching it up

[upper karma]

Possibly (No)

[fringe perch]

Hell nah

@oblique widget here u go

Got it?

[oblique widget]

Let me look

[upper karma]

Nope

[oblique widget]

Where did the 2 come
From

[fringe perch]

Ok

[fringe perch]

Which one?

[oblique widget]

The first step distributive property

[fringe perch]

You see 2l +2b= perimeter

[oblique widget]

Why is it l and b

[fringe perch]

Take the 2 out as common

[oblique widget]

I have different variables

[fringe perch]

Oh

[oblique widget]

You wrote wrong

[upper karma]

The idea is the same

He used different variables

[oblique widget]

I DONT UNDERSTAND

[fringe perch]

We use different variables here

[upper karma]

this problem is dumb

[fringe perch]

Yes

[upper karma]

I didn't solve for each individual variable

[fringe perch]

And i took 3 minutes to solve it

[oblique widget]

My teacher didn’t even teach this

[upper karma]

What i did was make x=(w+3)/2

But

I give up

Everytime I do the problem over, I get different values for x and y

[oblique widget]

AHH

ME TOO IM TRYING TO SOLVE FOR W TO PLUG IT

[upper karma]

W might say -11/9 so at this point I'm giving up

Best of luck

[oblique widget]

NOOOOOOOOO

@fringe perch help

[fringe perch]

Yeah

What is it?

@oblique widget

What is it?

[oblique widget]

Sorry

Help me with the problem

Like explain it verbally

@fringe perch

[fringe perch]

Verbally?

[oblique widget]

Like text

[fringe perch]

Ask someone else , I'm not good at English!

[fringe perch]

Wait then

[oblique widget]

Don’t write it on paper just like tell me step by step

[fringe perch]

Gimme 5 ninutes il

[oblique widget]

And I write as u say

[fringe perch]

Can I get the digital copy of that question pls!

@oblique widget where have you gone?

[oblique widget]

Uh whag do you mean digital

[oblique widget]

I’m doing my work

[fringe perch]

I mean send me the question

Il explain u

[fringe perch]

First let's take ad as l

And ab as b

We know that 2(Ab+Ad) =22

Ab +ad =11?

@oblique widget

[oblique widget]

Ok

[fringe perch]

So you could apply the value of ab and ad in the equation we got

You'll get

2y+1+3x-2=11

Solve this

@oblique widget

Did you get it?

[oblique widget]

One sec

[fringe perch]

Make it fast

I should sleep

[oblique widget]

Oh sorry I can just do it myself then

[fringe perch]

Wait Ill finish it off and go

[fringe perch]

You'll get 2y+3x =12

[oblique widget]

o

Ok

[fringe perch]

Upon using ur brain

Just plug some random values and see if it's equal to 12

If u plug in x as 2 and y as 3

Youll get 12?

So we can conclude that y =3

Line ab =2y +1

=7

Any queries?

@oblique widget

[oblique widget]

Ok

[fringe perch]

Got it?

@oblique widget

Or any doubts u have?

[oblique widget]

I’m doing ab = cd

Then ad =cb

[faint pasture]

I tried but got an error

It said sonething like

Error 404 brain not found

[fringe perch]

You need medical attention

[faint pasture]

Fr

[fringe perch]

I got no brain too

How does it look like?

[faint pasture]

Its looks grey

[fringe perch]

How do u know if you don't have one?

[faint pasture]

Everything is grey

You grey me grey everything grey

[fringe perch]

Grey everywhere

[sour hearth]

What grade

[tiny tartan]

Hey friends is there any way I can complete straight lines chapter in one day?

I have my maths exam in 10 days and I am cooked

• Basic concepts of Points and their
coordinates.
• The straight line

• Slope or gradient of a line.
• Angle between two lines.
• Condition of perpendicularity and
  parallelism.
• Various forms of equation of lines.
• Slope intercept form.
• Two-point slope form.
• Intercept form.
• Perpendicular /normal form.
• General equation of a line.
• Distance of a point from a line.
• Distance between parallel lines.
• Equation of lines bisecting the angle
  between two lines.
• Equation of family of lines
• Definition of a locus.
• Equation of a locus.

This is the syllabus

[graceful talon]

Yes :D

[graceful talon]

Are you starting from the beginning or you have previous/some knowledge of it?

[raw slate]

Morning (3-4 hours):
Basics First
Points and Coordinates: Just do a quick brush-up here, maybe 30 mins tops.
The Straight Line:
Slope/Gradient (30 mins) – get how to calculate and why it matters.
Angle Between Two Lines (30 mins) – learn the formula and maybe run through a few examples.
Perpendicularity and Parallelism (30 mins) – focus on those conditions.
Line Equations (1-1.5 hours):
Cover all the forms: slope-intercept, two-point, intercept, perpendicular/normal, and general form. Make sure you can convert between them smoothly.
Afternoon (3-4 hours): Application TimeDistance Problems (1.5 hours):
Point to Line Distance (45 mins) – know the formula and practice.
Distance Between Parallel Lines (45 mins) – same deal, understand then practice.
Advanced Line Stuff (1.5 hours):
Equation for Lines Bisecting Angles (45 mins) – get the concept and practice.
Family of Lines (45 mins) – focus on deriving and using the equations.
Evening (2-3 hours):
Locus and Review
Locus Problems (1.5 hours):
Definition + Equation (1 hour) – dive into understanding and deriving equations.
Quick Review + Practice (30 mins) – hit the key points again and do a few problems.
Final Hour: Recap & Test urself:
Go over formulas and concepts quickly. Do some mixed problems to see where you're at.

[raw slate]

Well.. i indeed have nothing to do if i wrote all of this

[fringe perch]

bro prepared a time table

[raw slate]

[thorn aspen]

I am NOT spending an hour and a half on line equations

[dense smelt]

How are lines m and l parallel

[obsidian harness]

use the alternate segment theorem

hence AC is the diameter of the circle

the diameter makes an angle of 90 degrees to both of the tangents

hence proven

by angles in the same segment, it doesn't matter where point D is

[dense smelt]

Ok I must have forgotten some properties of parallel lines

[obsidian harness]

no you forgot your circle theorems

[dense smelt]

How does the diameter making 90 w the lines tell us there parallel

And wasn’t that the diameter making 90,s alr given

[obsidian harness]

so if line l is perpendicular to the diameter, and m is also perpendicular to the diameter
lines l and m must be parallel

[dense smelt]

Or can we not assume that

[obsidian harness]

no you need to show that

I mean if the angles were different, you could have something like

[dense smelt]

Ig this would make sense for like alternate interior and all that

[obsidian harness]

actually there's an even simpler way that I overlooked cause I thought it was assuming the q

[dense smelt]

Dang I gotta stop assuming stuff

[obsidian harness]

draw a parallel line to line l through point B

oh wait I think that's assuming what we want to prove

[dense smelt]

I was trying to make a line through b and show that l and m are parallel to that so then l and m are parallel but it didn’t work

[obsidian harness]

yeah I think you actually need this

[obsidian harness]

yeah I thought of that too, draw a perpendicular line to line l through point B

the problem is that you can't show that the two perpendiculars coincide

so you haven't shown you have a straight line through B

[dense smelt]

Wow

It’s so easy to assume in geometry

Thanks for explaining this tho

[obsidian harness]

btw I found AOPS's solution

[obsidian harness]

no worries

[dense smelt]

Yea I have the solution manual you just made me realize tho that the AC is the transversal and that we had to prove it’s a str8 line

[empty yew]

How to prove $\sin^{2}A + \sin^{2}B - \sin^{2}C = 2 \sin{A}\sin{B}\cos{C}$

[somber coyote]

KingDanger

[empty yew]

Hi @obsidian harness After long time

My work

[obsidian harness]

https://approach0.xyz/search/?q=OR content%3A%24\sin^{2}A %2B \sin^{2}B - \sin^{2}C %3D 2 \sin{A}\sin{B}\cos{C}%24&p=1

approach0 is goated

hah I was thinking of the circumscribed radius approach

and someone answered, slick proof

[faint hollow]

Why the hell did you not write the given condition

it felt like something was missing

[obsidian harness]

it's kind of obvious if you've seen enough of these

that ABC is a triangel

[faint hollow]

I know, I have proved it in the past

But you never assume

[empty yew]

What is circumscribed radius approach?

[obsidian harness]

https://external-content.duckduckgo.com/iu/?u=https%3A%2F%2Fi.ytimg.com%2Fvi%2Fl6ol0gJS44Q%2Fmaxresdefault.jpg&f=1&nofb=1&ipt=9f0f836d7ab8737eacf0ac7ecf6df46832ddcf7f28d8b4b16351ce17aa8ab3d1&ipo=images

so $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}$

[somber coyote]

south

[obsidian harness]

oh circumradius I can't English

[empty yew]

Sry

I have the topic law of sines and cosines in my later topics but now I want to solve this with sum or product and conditions

[tiny tartan]

Previous knowledge I have

[steady vault]

ez

[odd horizon]

how do you get the first step

[faint hollow]

They are half angle properties taught in trig

[faint pasture]

Do you want like the derivation?

Cuz its pretty doable use sin4x = 2 sin2x cos2x

Now try to express this in terms of only tan

Similarly you can do cos

[odd horizon]

yeah

[odd horizon]

im stuck with a stray cos^2 2a at some random point

[faint pasture]

Write it as 1/sec^2 2a

[odd horizon]

hmm

OH

i see

i forgot

oops

thanks

[quartz minnow]

I need help with the one marked with H

[hoary prism]

Basically they gave you a circle and and two legs of an inscribed triangle and they’re asking you to find the third

[quartz minnow]

Yep I did it

[hoary prism]

Nice

[hallow ridge]

does anyone know what the Sc on question e represent?

[grave mulch]

I need help with this question, your supposed to use this format to find the answer: y=ax+b

[verbal tapir]

so, first we know one point is on (-1,2) and the other on (3,5)

the slope of a line that connects these two points are (5-2)/(3-(-1))

=3/4

thus, a, which represents the slope, is 3/4

to find the b, we can replace y with (y-y_1) and x with (x-x_1)
(y_1 and x_1 are the coordinates of one of the two given points, although any point on the line works)

for this solution, i'll use (3,5)

so, y-5=(3/4)(x-3)

expand this to get
y-5=(3/4)x-9/4
add 5 on both sides to get the final equation
y=(3/4)x+11/4

i should learn how to use the latex bot

[grave mulch]

i understand now thanks

but i use this method which is very similar to yours

[verbal tapir]

yeah that also works

as far as i know, there are a lot of methods but all of them involve using the two points to find the slope and then one to find the y intercept

(at least all the methods i know)

[prime stratus]

the number 5

[verbal tapir]

i don't believe this goes here

but sure

in an arithmetic sequence, given a term aₙ (where n is not 1), aₙ₋₁=aₙ-d and aₙ₊₁=aₙ+d

aₙ₋₁+aₙ₊₁ = 2aₙ-d+d = 2aₙ

Thus, 7x-1+6x-9 = 2*8 = 16
13x-10=16
13x=26, x=2

Therefore,
8x+2=18
7x-1=13
8
6x-9=3

[verbal tapir]

late reply (and thus ping)
but i believe S represents the sample space (i. e. the set of all possible otucomes)

[obsidian harness]

<@&268886789983436800> spam link

[surreal void]

Hello

[full nymph]

hi

[hallow ridge]

thank you

[kind quail]

Hello

[tribal rose]

How do you reflect shapes over a diagonal line?

[gusty spruce]

Isn’t it just sequence a

[tribal rose]

If you rotate it 270 degrees, isn't Point I to the right of J?

[gusty spruce]

Yeah your right it’s sequence b

I was confused on where the origin was

[tribal rose]

How did you reflect the shape over HI?

I'm confused as to how to do diagonal reflections.

[gusty spruce]

I just did the individual points

[tribal rose]

So like this (roughly)?

[gusty spruce]

Yeah

I’m honestly a little confused so I just went to the points

Then the rest is self explanatory

[tribal rose]

It looks kind of like a 90-degree rotation.

[gusty spruce]

I saw it like that yeah

[verbal tapir]

just gonna say this

the reflection does seem similar to a rotation, but without some sort of symmetry, it is impossible to have two congruent shapes be able to overlap perfectly using only rotation if one is a reflection of the other

just to clarify

[unique nacelle]

cannot be 90 degree

[tribal rose]

So it's not really a rotation.

I'm confused as to how this works, then.

[unique nacelle]

NOO

LOOK!

ITS REFLECTION

MAKE THE LINE PERPENDICUILAr

THEN SAME DISTENCE ON THE OTEHR SIDE

[verbal tapir]

let's reflect a point with respect to a line

first, we find the foot of perpendicular from P to l
the distance is d

extend the line segment by d

the end point is P' (the reflection of P)

now, other shapes are just a set of points

shapes can be reflected like this

for polygons, or angled straight lines, reflect each vertex and then connect the corresponding points

for other shapes like circles which only require a little information, reflect specific necessary points (like the center of a circle) and use the information from earlier to draw the shape

[verbal tapir]

applying it here,

(assuming the side length of a grid square is 1)

[tribal rose]

Thank you so much! This was really clear.

So it's just two points that are equidistant from each other.

[oblique widget]

Somebody like explain this to me

[verbal tapir]

each color represents the different possible parallelograms

[twin spindle]

Im not tripping right? This is correct?

[verbal tapir]

yeah that seems correct

(n is an integer)

[twin spindle]

Brooo, why'd my teacher doc me 😭😭

[oblique widget]

what the fart why did nobody help me

[verbal tapir]

the graph shows the three possible pointsfor the triangle to be half of a parallelogram

[twin spindle]

How abt problems like these? Cause doesn't the sin and sin-1 cancel?

Which jus leaves the degree

[verbal tapir]

no

keep in mind the range of sin^-1

[twin spindle]

Yeah -pi/2<ø<pi/2

[verbal tapir]

because sin is a periodic function, different angles can have the same sine value

[twin spindle]

So it would jus be sqrt(2)/2

[verbal tapir]

-225 degrees is smaller than -pi/2 radians

which means the value of a sin^-1 function can never be -225 degrees

(or its equivalent in radians)

[twin spindle]

Yeah so we would have to find a degree that's the same value as -225⁰ and between this interval

Right?

[verbal tapir]

yes exactly

[twin spindle]

Which would be -60⁰

[verbal tapir]

its

-45 actually

but yeah that is the idea

[twin spindle]

Wait your right mb mb 😭😭

[verbal tapir]

dw that happens

[twin spindle]

There was one like this to that said cos(cos-1 (5))

But that would be undefined, right?

[verbal tapir]

5?

[twin spindle]

Cause a value has to be less then 1 if you were to take the cos-1 right?

Yeah

[verbal tapir]

yeah i think that's undefined

[twin spindle]

Yeah I got it wrong cause at the time I thought the cos would cancel

But I woulda had to say "undefined"

[verbal tapir]

ohh ok

yeah be mindful of these 'traps'

[twin spindle]

Yeah I fell for hella on this section

Like on this one, tbh I had no clue what to do, but I saw cos/sin and immediately thought of Cot(10) 😭

[verbal tapir]

oh

[twin spindle]

But I think you were supposed to apply the complementary angle formula

[verbal tapir]

yeah

that gives 1

[twin spindle]

Idk when to apply those tbh

[verbal tapir]

when you are given two values to operate with

if the two values add up to 90+360n degrees (or 270+360n) you can assume you might need to use it later on

(i believe you do have to take the signs in account for 270)

[twin spindle]

I see I see

[verbal tapir]

sin(θ)=cos(90-θ)
sin(θ)=-cos(270-θ)

[twin spindle]

I thought this problem you'd have to apply the complementary formula 😭

[verbal tapir]

oh

[twin spindle]

Or I think you do, but I somehow forgot abt the sin

[verbal tapir]

the last question?(question g)

[twin spindle]

The one is cos(90-22.5)

[verbal tapir]

oh

[twin spindle]

Yeah

[verbal tapir]

that one doesn't 'work' because neither 22.5 nor 67.5 are the special angles

[twin spindle]

Oh wrrd they have to be a special angle 😭

[verbal tapir]

you have to use the half-angle formula of cosine from cos(45)

either that or

draw a triangle

[twin spindle]

What 😭😭

How do yk when to apply thaf 💀💀

[verbal tapir]

when you get angles that aren't 0, 30, 45, 60, 90

(or +90n variants of them)

you have a few options

if the target angle can be split into 2 of the special angles
such as 75 degrees (30+45), you use the addition formula

if the target angle is half of a special angle such as 15 degrees or 22.5 degrees (half of 30 and 45, respectively), you use the half angle formula

of course, this only applies to numerical values of the angle, if it's given as a variable like θ, you have to use a different approach for each scenario

when learning about formulas and such, it's a good idea to learn or think about what scenario you want to use them in

[twin spindle]

Shi I see I see

I didn't think this test would have us apply any sort of formulas on this section

Thought it would jus be common angles and finding the values 💀💀

So is this how you would show the work for this problem?

[verbal tapir]

tbh i don't know your curriculum

but id say yes

[twin spindle]

This means that we would have had to apply the sum or difference formula

[verbal tapir]

yes

exactly

[twin spindle]

On this question

Omg

I'm pissed 😭😭

I sold this test

Their were other problems that I screwed up on aswell, but that's because I wrote the wrong values from the unit circle 😭

[verbal tapir]

oh that's rough

fortunately, you won't be messing these up in the future now

[twin spindle]

Oh yeah fs

I knew I shoulda wrote the values on the unit circle that my teacher gave us 😭

Should i ask if I could retake 💀💀

Couldn't hurt to ask, right?

[verbal tapir]

don't know about that myself tbh

i only know geometry and a bunch of other miscellaneous stuff

[twin spindle]

Fair

[verbal tapir]

but if you think you should, go for it

[twin spindle]

Alr I might. I got a couple more, How abt these couple of questions?

[verbal tapir]

i don't get how you did the first question

considering the value of β isn't shown

now, i don't know exactly what the paper means by 'direction'
but the magnitude is just pythagoras applied to 7 and 3

for question 5a, that's just arithmetic

if the direction means the direction relative to the positive x axis, the direction of the Q4 would just be tan^-1(3/7)

the angle between the Q5 vectors u and v is given by the formula
cosθ=(u⋅v)/(|u||v|)

where u⋅v is the inner product of u and v

[twin spindle]

Oh yeah mb

Their was multiple parts in this section, with using the same cos and sin

[twin spindle]

Yeah I realized that after relooking it. But direction is finding the angle which at the time I think I didn't for some didn't see that portion of the question, and didnt do it

[twin spindle]

Oh shi, so the dot product?

[verbal tapir]

yeah

[twin spindle]

I thought all you had to apply was tan

[verbal tapir]

hmm

yeah

you can make it work with tan too

[twin spindle]

But you would have to add 180 right?

Since its in the 2nd quadrant

[obsidian harness]

tan theta = y/x only works when you have one vector and you want to find the angle with the x-axis

the intended method is to find cos theta using the dot product then use the Pythagorean identity / draw a right triangle to find tan theta

[verbal tapir]

seeing that the angles are approximate (at least from what i can see)
i think just getting two tan^-1 values and subtracting them should work

[obsidian harness]

there are definitely other methods too

[twin spindle]

But since we're working w 2 vectors it doesn't work

[obsidian harness]

if you draw it out, you have that v is in the first quadrant and u is in the fourth quadrant

so $\theta = \arctan \frac{2}{3} + \arctan \frac{3}{4}$

$\tan \theta = \frac{2/3 + 3/4}{1 - (2/3)(3/4)}$

[obsidian harness]

it works if you think more creatively

[verbal tapir]

answer to question b is wrong, the value you got is higher than 1

[pearl bolt]

root3 COT^2 - 4COTA + root 3

Guys please help me in solving theis

this

[obsidian harness]

oh wait it's 3/4

[somber coyote]

south

[pearl bolt]

root3 COT^2 - 4COTA + root 3

Guys please help me in solving theis
this

[obsidian harness]

stop spamming

also you want this equal to 0 right?

[pearl bolt]

No, COT^2 + TAN^2 = ?

[obsidian harness]

there's no identity for that

[obsidian harness]

you also didn't answer my question regarding this

[pearl bolt]

root3 COT^2 - 4COTA + root 3

then COT^2 + TAN^2 = ?

Guys please help me in solving theis
this