If tan theta cdot tan 2 theta = 1, then the v | vedclass

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(C) Given $	an 	heta \cdot 	an 2 	heta = 1.$This implies $	an 	heta = \frac{1}{	an 2 	heta} = \cot 2 	heta.$We can write $\cot 2 	heta$ as $

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$3 \frac{3}{4}$

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(C) Given $	an 	heta \cdot 	an 2 	heta = 1.$This implies $	an 	heta = \frac{1}{	an 2 	heta} = \cot 2 	heta.$We can write $\cot 2 	heta$ as $	an(90^{\circ} - 2 	heta).$So,$	an 	heta = 	an(90^{\circ} - 2 	heta).$Equating the angles,$	heta = 90^{\circ} - 2 	heta \Rightarrow 3 	heta = 90^{\circ} \Rightarrow 	heta = 30^{\circ}.$Now,substitute $	heta = 30^{\circ}$ into the expression $\sin ^{2} 2 	heta + 	an ^{2} 2 	heta.$This becomes $\sin ^{2} 60^{\circ} + 	an ^{2} 60^{\circ}.$Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$ and $	an 60^{\circ} = \sqrt{3},$we get $(\frac{\sqrt{3}}{2})^{2} + (\sqrt{3})^{2} = \frac{3}{4} + 3 = 3 \frac{3}{4}.$

If $\tan \theta \cdot \tan 2 \theta = 1,$ then the value of $\sin ^{2} 2 \theta + \tan ^{2} 2 \theta$ is equal to

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