Question

True or false. Tan^2 x = 1 - cos2x/ 1 + cos 2x

Answer 1

ANSWER

True

EXPLANATION

The given trigonometric equation is

$\tan^{2} (x) = \frac{1 - \cos(2x) }{1 + \cos(2x) }$

Recall the double angle identity:

$\cos(2x) = \cos^{2} x - \sin^{2}x$

We apply this identity to obtain:

$\tan^{2} (x) = \frac{1 - (\cos^{2} x - \sin^{2}x) }{1 + (\cos^{2} x - \sin^{2}x) }$

We maintain the LHS and simplify the RHS to see whether they are equal.

Expand the parenthesis

$\tan^{2} (x) = \frac{1 - \cos^{2} x + \sin^{2}x }{1 + \cos^{2} x - \sin^{2}x}$

$\implies\tan^{2} (x) = \frac{1 - \cos^{2} x + \sin^{2}x }{1 - \sin^{2}x + \cos^{2} x }$

Recall that:

$1 - \sin^{2}x = \cos^{2}x$

$1 - \cos^{2}x = \sin^{2}x$

We apply these identities to get:

$\implies\tan^{2} (x) = \frac{\sin^{2}x + \sin^{2}x }{\cos^{2} x + \cos^{2} x }$

$\implies\tan^{2} (x) = \frac{2\sin^{2}x }{ 2\cos^{2} x }$

$\implies\tan^{2} (x) = \frac{\sin^{2}x }{ \cos^{2} x }$

$\implies \tan^{2} (x) =( \frac{\sin x }{ \cos x })^{2}$

Also

$\frac{\sin x }{ \cos x } = \tan(x)$

$\implies \tan^{2} (x) =( \tan x )^{2}$

$\implies \tan^{2} (x) =\tan^{2} (x)$

Therefore the correct answer is True