Question

Can somebody help me solve this please? Thanks in advance!

`(tan^2x-1)/sec^2x=(tanx-cotx)/(tanx+cotx)`

Answer 1

See explanation

Explanation:

We want to verify the identity

`(tan^2(x)-1)/sec^2(x)=(tan(x)-cot(x))/(tan(x)+cot(x))`

We will use the pythagorean trigonometric identity

• `sin^2(x)+cos^2(x)=1`

And some definitions of trigonometric functions

`RHS=(tan(x)-cot(x))/(tan(x)+cot(x))`

`=(sin(x)/cos(x)-cos(x)/sin(x))/(sin(x)/cos(x)+cos(x)/sin(x))`

`=((sin^2(x)-cos^2(x))/(cos(x)sin(x)))/((sin^2(x)+cos^2(x))/(cos(x)sin(x))`

`=(sin^2(x)-cos^2(x))/(sin^2(x)+cos^2(x))`

`=(sin^2(x)-cos^2(x))/1`

`=((sin^2(x)-cos^2(x))sec^2(x))/sec^2`

`=(sin^2(x)/cos^2(x)-cos^2(x)/cos^2(x))/sec^2`

`=(tan^2(x)-1)/sec^2(x)=LHS`