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

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(A) Given: $\frac{1 - 	an^2 	heta}{\sec^2 	heta} = \frac{1}{2}$ Since $\frac{1}{\sec^2 	heta} = \cos^2 	heta$ and $	an^2 	heta = \frac{\sin^2 \

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

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(A) Given: $\frac{1 - 	an^2 	heta}{\sec^2 	heta} = \frac{1}{2}$ Since $\frac{1}{\sec^2 	heta} = \cos^2 	heta$ and $	an^2 	heta = \frac{\sin^2 	heta}{\cos^2 	heta}$,we have: $\cos^2 	heta (1 - \frac{\sin^2 	heta}{\cos^2 	heta}) = \frac{1}{2}$ $\cos^2 	heta - \sin^2 	heta = \frac{1}{2}$ Using the identity $\cos 2	heta = \cos^2 	heta - \sin^2 	heta$,we get: $\cos 2	heta = \frac{1}{2} = \cos(\frac{\pi}{3})$ The general solution for $\cos x = \cos \alpha$ is $x = 2n\pi \pm \alpha$. Therefore,$2	heta = 2n\pi \pm \frac{\pi}{3}$ Dividing by $2$,we get $	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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