Question

Verify each identity:

Answer 1

\[\frac{ \cos^2x-\sin^2x }{ 1-\tan^2x }=\cos^2\]

Answer 2

\(\bf \cfrac{cos^2(x)-sin^2(x)}{1-tan^2(x)}\implies \cfrac{cos^2(x)-sin^2(x)}{1-\frac{sin^2(x)}{cos^2(x)}}\implies \cfrac{cos^2(x)-sin^2(x)}{\frac{\square -\square }{{\color{red}{ LCD}}}}\)

add the denominator, see what you get?

Answer 3

\[\frac{ \cos^2x-\sin^2x }{ \frac{ \cos^2x-\sin^2x }{ \cos^2x } }\]

Answer 4

\(\bf \huge {\frac{ \cos^2x-\sin^2x }{ \frac{ \cos^2x-\sin^2x }{ \cos^2x } }\implies \frac{ \cancel{\cos^2x-\sin^2x} }{ 1 }\cdot \frac{ \cos^2x }{ \cancel{\cos^2x-\sin^2x }}}\)

Answer 5

ohhhhhhh :) Thanks!!! :)