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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $\tan^2\theta=2\tan^2\phi+1$ then $\cos 2\theta+\sin^2\phi$ is equal to

$(a)\;1\qquad(b)\;2\qquad(c)\;-1\qquad(d)None\;of\;these$

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1 Answer

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We have
$\cos 2\theta=\large\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$\qquad\quad=\large\frac{1-(2\tan^2\phi+1)}{1+2\tan^2\phi+1}$
$\tan^2\theta=2\tan^2\phi+1$
$\qquad\quad=\large\frac{-2\tan^2\phi}{2+2\tan^2\phi}$
$\qquad\quad=\large\frac{-\tan^2\phi}{\sec^2\phi}$
$\qquad\quad=-\sin^2\phi$
$\therefore \cos2\theta+\sin^2\phi=0$
Hence (d) is the correct answer.
answered Oct 4, 2013 by sreemathi.v
 

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