If frac{{1 - {{tan }^2}theta }}{{{{sec }^2}th | vedclass

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Question

(a) $\frac{{1 - {{	an }^2}	heta }}{{{{\sec }^2}	heta }} = \frac{1}{2} $

$\Rightarrow {\cos ^2}	heta - {\sin ^2}	heta = \frac{1}{2}$

$ \Righ

Vedclass · Accepted Answer

$n\pi \pm \frac{\pi }{6}$

Vedclass · Answer

(a) $\frac{{1 - {{	an }^2}	heta }}{{{{\sec }^2}	heta }} = \frac{1}{2} $

$\Rightarrow {\cos ^2}	heta - {\sin ^2}	heta = \frac{1}{2}$

$ \Rightarrow $ $\cos 2	heta = \frac{1}{2} = \cos \left( {\frac{\pi }{3}} ight)$

$ \Rightarrow $ $2	heta = 2n\pi \pm \frac{\pi }{3} $

$\Rightarrow 	heta = n\pi \pm \frac{\pi }{6}$.

If $\frac{{1 - {{\tan }^2}\theta }}{{{{\sec }^2}\theta }} = \frac{1}{2}$, then the general value of $\theta $ is

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