[SPLIT] prove (csc^2x-cot^2x)/(1+tan^2x)=cos^2x

[Sham]

Hello, Sham!

Do you know any of the basic identities?

. . \(\displaystyle \cos x\,=\,\frac{1}{\sec x}\;\;\;\sec x \,=\,\frac{1}{\cos x}\)

. . \(\displaystyle \sin x \,=\,\frac{1}{\csc x}\;\;\;\csc x\,=\,\frac{1}{\sin x}\)

. . \(\displaystyle \tan x\,=\,\frac{\sin x}{\cos x}\;\;\;\cot x\,=\,\frac{\cos x}{\sin x}\)

ok the rules which go for the first two problems (from the other thread) also go for this problem :

csc^2 x - cot^2 x
----------------------- = cos^2 x
. . .1 + tan^2 x

 

[stapel]

ok the rules which go for the first two problems (in the other thread) also go for this problem :

csc^2 x - cot^2 x
--------------------- = cos^2 x
1 + tan^2 x

How did your class get all the way to proving identities without ever actually covering the basic trig functions? :shock:

Yes, the definitions of the trig functions are always the same. You will also need the Pythagorean Identity, cos²(x) + sin²(x) = 1.

Eliz.

 

[skeeter]

\(\displaystyle \L \frac{\csc^2{x} - \cot^2{x}}{1 + \tan^2{x}} = \cos^2{x}\)

while you're at it, you might as well learn the three Pythagorean identities ...

\(\displaystyle \L \cos^2{x} + \sin^2{x} = 1\)
\(\displaystyle \L 1 + \tan^2{x} = \sec^2{x}\)
\(\displaystyle \L 1 + \cot^2{x} = \csc^2{x}\)

use these to prove your identity.