If {tan ^2}theta = 2{tan ^2}phi + 1, then cos | vedclass

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(b) ${	an ^2}	heta = 2{	an ^2}\phi + 1 $

$\Rightarrow 1 + {	an ^2}	heta = 2\,(1 + {	an ^2}\phi )$ 

==> ${\sec ^2}	heta

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(b) ${	an ^2}	heta = 2{	an ^2}\phi + 1 $

$\Rightarrow 1 + {	an ^2}	heta = 2\,(1 + {	an ^2}\phi )$ 

==> ${\sec ^2}	heta = 2{\sec ^2}\phi $

$\Rightarrow {\cos ^2}\phi = 2{\cos ^2}	heta $ 

==> ${\cos ^2}\phi = 1 + \cos 2	heta $

$\Rightarrow {\sin ^2}\phi + \cos 2	heta = 0$.

Trick : Let $	heta = {45^o}$, then $\phi = 0$

$	herefore \;\cos (2 	imes {45^o}) + {\sin ^2}0 = 0 + 0 = 0$.

If ${\tan ^2}\theta = 2{\tan ^2}\phi + 1,$ then $\cos 2\theta + {\sin ^2}\phi $ equals

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