Question

Trigonometry

Answer 1

Put tan x = sin x / cos x

Answer 2

\(\large \dfrac{\cos^2 x - \sin^2 x}{1-\tan^2 x} \)

\(=\large \dfrac{\cos^2 x - \sin^2 x}{1-\frac{sin^2 x}{\cos^2 x}} \)

\(=\large \dfrac{\cos^2 x - \sin^2 x}{\frac{cos^2 x}{\cos^2 x}-\frac{sin^2 x}{\cos^2 x}} \)

Now you can finish it.

Answer 3

You should get \(1-\tan ^2 x = \frac{cos^2 x - sin^2 x}{cos^2 x}\)

Answer 4

thank you everyone! I got it :)
answer is cos^2 x?

Answer 5

Correct.

Answer 6

Great, thank you so much @ganeshie8
@mathstudent55
@ShailKumar
@zepdrix

Answer 7

You're welcome.