The identity [tex]\tan^2x = \frac{1 -\cos 2x}{1 + \cos 2x}[/tex] is true
How to prove or disprove the identity?
The identity is given as:
[tex]\tan^2x = \frac{1 -\cos 2x}{1 + \cos 2x}[/tex]
As a general rule;
[tex]\cos 2x = 1 - 2\sin^2 x = 2\cos^2x - 1[/tex]
So, we have:
[tex]\tan^2x = \frac{1 -(1 - 2\sin^2 x)}{1 + (2\cos^2x - 1)}[/tex]
Remove the brackets
[tex]\tan^2x = \frac{1 -1 + 2\sin^2 x}{1 + 2\cos^2x - 1}[/tex]
Evaluate the like terms
[tex]\tan^2x = \frac{2\sin^2 x}{2\cos^2x}[/tex]
Divide 2 by 2
[tex]\tan^2x = \frac{\sin^2 x}{\cos^2x}[/tex]
The above equation is true because:
[tex]\tanx = \frac{\sin x}{\cosx}[/tex]
Hence, the identity [tex]\tan^2x = \frac{1 -\cos 2x}{1 + \cos 2x}[/tex] is true
Read more about trigonometry identity at:
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