Question

Solve:

$\frac{1-ta{n}^2(\frac{\pi }{4}+\theta )}{1+ta{n}^2(\frac{\pi }{4}+\theta )}=$

• A. $sin2\theta$
• B. $-sin2\theta$
• C. $cos2\theta$
• D. $-cot2\theta$

Answer 1

The correct option is A $sin2\theta$

$\frac{1-{\tan }^2(\frac{\pi }{4}+\theta )}{1+{\tan }^2(\frac{\pi }{4}+\theta )}$

We know that $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$\frac{1-{\tan }^2(\frac{\pi }{4}+\theta )}{1+{\tan }^2(\frac{\pi }{4}+\theta )}=\cos 2(\frac{\pi }{4}+\theta )$

$=\cos (\frac{\pi }{2}+2\theta )$

$=-\sin 2\theta$