Trigonometry Proofs

[vocal yoke]

what are you trying to prove?

[frozen falcon]

that 1 + tan²(-θ) = sec²θ

[vocal yoke]

also cos²(θ) - sin²(θ) = cos²(2θ) shouldn't have a ^2 on the right side

This property is a corollary of sin^2 + cos^2 = 1

so I would start from there

[frozen falcon]

yeah but it's (cos²(θ) - sin²(θ)

since tan²(-θ)

[vocal yoke]

oh I see

well the property is still useful

$\sin^2(-\theta) + \cos^2(-\theta) = 1$

[quasi walrus]

Ari

[frozen falcon]

wait what but wouldnt it be -sin²(θ)

wait

nvm

-sin²(θ) = sin²(-θ) = (sin(-θ))² = sin²(θ)

???

OHHHH

THAT MAKE SENSE

THANK YOU @vocal yoke

[vocal yoke]

[frozen falcon]

any tips

that you could give me please

[vocal yoke]

Well your goal is to transform this into the identity you're trying to prove

do you see how you can perhaps change sin to tan

[frozen falcon]

nah

my brain is slow

i mean sine alone cant change to tangent right?

[vocal yoke]

remember that tan = sin/cos

[frozen falcon]

yes

but without cosine it can't be equal to tangent

so sin -= to tan

[vocal yoke]

What happens when you divide both sides of this by $\cos^2(-\theta)$?

[quasi walrus]

Ari

[frozen falcon]

that

that'll turn to sin(-x) + 1 = 1/cos^2x

[vocal yoke]

You get $\tan^2(-\theta) + 1 = \frac{1}{\cos^2(-\theta)}$

[quasi walrus]

Ari

[frozen falcon]

divide the whole left side with cos of theta my bad yes that one

but what r u trying to point out

[vocal yoke]

well what's another way to write the right side

i.e what's another way to write 1/cos

[frozen falcon]

oh

I see it

that another approach

but what if

like im stuck

let say

cos[(cosxsinx/sinxcosx) + *(cosxsinx/sinxcosx)] = csc

I started from tanx + cotx

I convert everything and multiply the common denominator

how do I know if my approach is right

[vocal yoke]

uh could you clarify this

[vocal yoke]

Well if you end up with the identity you're trying to prove then it's probably correct

[frozen falcon]

like so conevrt tan and cot to
cos[(sinx/cosx) + (cosx/sinx)]
cos[(cos(x)(sin(x))/cosx(sin(x))] + [(cos(x)(sin((x)/sin(x)(cosx)] = csc