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^2

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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: $\cos^2 	heta = \frac{1}{2} \cos^2 \phi$Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$:$\frac{1 + \cos 2	heta}{2} = \frac{1}{2} \cos^2 \phi$$1 + \cos 2	heta = \cos^2 \phi$$1 + \cos 2	heta = 1 - \sin^2 \phi$$\cos 2	heta + \sin^2 \phi = 0$Alternatively,let $	heta = 45^\circ$. Then $	an^2 45^\circ = 2	an^2 \phi + 1 \Rightarrow 1 = 2	an^2 \phi + 1 \Rightarrow 	an^2 \phi = 0 \Rightarrow \phi = 0^\circ$.Thus,$\cos(2 	imes 45^\circ) + \sin^2 0^\circ = \cos 90^\circ + 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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