Question

LS=RS
sec^2x - sec^2y = tan^2x-tan^2y

Answer 1

\[\cos^2(x)+\sin^2(x)=1 \implies 1 + \tan^2(x)=\sec^2(x)\]
\[\implies \boxed{\sec^2(x)-\sec^2(y)}=1+\tan^2(x)-1-\tan^2(y)=\boxed{\tan^2(x)-\tan^2(y)}\]

Answer 2

I have no idea what you did there

Answer 3

Used the pythagorean identity. Divide by cos^2(x) to rewrite it as the right side of the first equation. Then plug in what you have and cancel the 1s.

Answer 4

What is the pythagorean identity?

Answer 5

\[\boxed{\cos^2(x) + \sin^2(x)=1}\]

Answer 6

How does that apply within sec^2x - sec^2y = tan^2x-tan^2y ?

Answer 7

What is secant in terms of cosine?

Answer 8

secx = 1/cosx?

Answer 9

Exactly. So divide both sides of the pythagorean identity by cos^2(x):
\[\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} \implies 1 + \frac{\sin^2(x)}{\cos^2(x)} = \sec^2(x)\]
But what is sin(x)/cos(x)?

Answer 10

sinx/cosx = tan x?

Answer 11

Yes. So you have:
\[\sin^2(x)+\cos^2(x) = 1 \implies 1+\tan^2(x) = \sec^2(x)\]
And you have two sec^2(x) quantities. So replace those with this identity:
\[(1+\tan^2(x))-(1+\tan^2(y))=\tan^2(x)-\tan^2(y)\]
which is exactly what you wanted to show.

Answer 12

so sec^2x would = 1+tan^2x?

Answer 13

Yes.

Answer 14

So you would sub in 1 + tan ^2x/y into both sec^2x/y? what happens to the 1s?

Answer 15

1-1 = 0

Answer 16

Don't forget to distribute the negative.
\[ - (a+b) = -a-b\]

Answer 17

ah okay sorry brakets confused me

Answer 18

:)

Answer 19

oh thank you so much :)