ANSWER
True
EXPLANATION
The given trigonometric equation is
[tex] \tan^{2} (x) = \frac{1 - \cos(2x) }{1 + \cos(2x) } [/tex]
Recall the double angle identity:
[tex] \cos(2x) = \cos^{2} x - \sin^{2}x[/tex]
We apply this identity to obtain:
[tex]\tan^{2} (x) = \frac{1 - (\cos^{2} x - \sin^{2}x) }{1 + (\cos^{2} x - \sin^{2}x) } [/tex]
We maintain the LHS and simplify the RHS to see whether they are equal.
Expand the parenthesis
[tex]\tan^{2} (x) = \frac{1 - \cos^{2} x + \sin^{2}x }{1 + \cos^{2} x - \sin^{2}x} [/tex]
[tex] \implies\tan^{2} (x) = \frac{1 - \cos^{2} x + \sin^{2}x }{1 - \sin^{2}x + \cos^{2} x } [/tex]
Recall that:
[tex]1 - \sin^{2}x = \cos^{2}x [/tex]
[tex]1 - \cos^{2}x = \sin^{2}x[/tex]
We apply these identities to get:
[tex]\implies\tan^{2} (x) = \frac{\sin^{2}x + \sin^{2}x }{\cos^{2} x + \cos^{2} x } [/tex]
[tex]\implies\tan^{2} (x) = \frac{2\sin^{2}x }{ 2\cos^{2} x } [/tex]
[tex]\implies\tan^{2} (x) = \frac{\sin^{2}x }{ \cos^{2} x } [/tex]
[tex]\implies \tan^{2} (x) =( \frac{\sin x }{ \cos x })^{2} [/tex]
Also
[tex] \frac{\sin x }{ \cos x } = \tan(x) [/tex]
[tex]\implies \tan^{2} (x) =( \tan x )^{2} [/tex]
[tex]\implies \tan^{2} (x) =\tan^{2} (x) [/tex]
Therefore the correct answer is True
