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

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(B) Given: $	an^2 	heta = 2	an^2 \phi + 1$ Adding $1$ to both sides: $1 + 	an^2 	heta = 2	an^2 \phi + 2$ $\Rightarrow \sec^2 	heta = 2(1 + 	an

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(B) Given: $	an^2 	heta = 2	an^2 \phi + 1$ Adding $1$ to both sides: $1 + 	an^2 	heta = 2	an^2 \phi + 2$ $\Rightarrow \sec^2 	heta = 2(1 + 	an^2 \phi)$ $\Rightarrow \sec^2 	heta = 2\sec^2 \phi$ Taking the reciprocal: $\cos^2 	heta = \frac{1}{2} \cos^2 \phi$ $\Rightarrow 2\cos^2 	heta = \cos^2 \phi$ Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we have $2\cos^2 	heta = 1 + \cos 2	heta$ Also,$\cos^2 \phi = 1 - \sin^2 \phi$ Substituting these into the equation $2\cos^2 	heta = \cos^2 \phi$: $1 + \cos 2	heta = 1 - \sin^2 \phi$ $\Rightarrow \cos 2	heta + \sin^2 \phi = 0$.

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

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