Question

If $ta{n}^2\theta =2ta{n}^2\varphi +1$,then $cos2\theta +si{n}^2\varphi =$

• A. $0$
• B. $1$
• C. $2$q
• D. None of these

Answer 1

The correct option is A $0$

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

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

We have ${\tan }^2\theta =2{\tan }^2\varphi +1$

$=\frac{-2{\tan }^2\varphi }{2{\sec }^2\varphi }$

$=\frac{-{\tan }^2\varphi }{{\sec }^2\varphi }$

$=-{\tan }^2\varphi {\cos }^2\varphi$

$=-\frac{{\sin }^2\varphi }{{\cos }^2\varphi }\times {\cos }^2\varphi$

$=-{\sin }^2\varphi$

$\therefore \cos 2\theta +{\sin }^2\varphi =0$

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