can someone explain this to me plz!

[halcyon karma]

@timber plover i can help

what we want to do is flip the fraction so we have an easier trigonometric function to work with

,tex $\frac{1}{sec(\theta)} = \frac{1}{\frac{9}{8}}$

[low timber]

@halcyon karma

[halcyon karma]

,tex $cos(\theta) = \frac{8}{9}$

[low timber]

@halcyon karma

[timber plover]

Yea i did that, but after that what should i do

[halcyon karma]

we know that theta is in the 4th quadrant, which means that the X coordinate is positive, and the Y coordinate is negative

cosine always refers to the X coordinate divided by the radius of the circle

so if $r = 1$, then $x = \frac{8}{9}$

[low timber]

@halcyon karma

[halcyon karma]

@timber plover do u understand how i got this

[timber plover]

Yes i got it

[halcyon karma]

ok

now can you use the pythagorean theorem to solve for the Y-coordinate

[timber plover]

So i should just solve it without converting it to tan

[halcyon karma]

sry i dont understand what you mean

[timber plover]

Because the question want me to find (tan)

[halcyon karma]

yes, tan refers to the Y-coordinate divided by the X-coordinate

[halcyon karma]

we don't know the Y-coordinate yet, but we can find it using the pythagorean theorem

[timber plover]

Let me try

So x=a/b

Right?

[halcyon karma]

sorry?

[halcyon karma]

what do A and B refer to?

[timber plover]

I mean a= 8 and b=9

[halcyon karma]

yes

[timber plover]

Referring to x

[halcyon karma]

yes, a/b refers to the x-coordinate

or technically its a/c

,tex $a^{2} + b^{2} = c^{2}$

[low timber]

@halcyon karma

[halcyon karma]

,tex $(\frac{8}{9})^{2} + b^{2} = 1^{2}$

[low timber]

@halcyon karma

[timber plover]

Ohh wait

[halcyon karma]

youre close

but remember, X-coordinate is positive and Y-coordinate is ________

[timber plover]

Negative

[halcyon karma]

yep

[halcyon karma]

so how do we fix the answer than you got

[halcyon karma]

correct

,tex $tan(\theta) = \frac{y}{x}$

[low timber]

@halcyon karma

[halcyon karma]

,tex $tan(\theta) = \frac{- \frac{\sqrt[2]{17}}{9}}{\frac{8}{9}}$

[low timber]

@halcyon karma

[halcyon karma]

yep

[timber plover]

thank u soooooo much!!!

[olive forum]

or you could use tan^2x+1=sec^2x next time

[timber plover]

Thank u so much 💕💕