tan left(frac{3 pi}{16}right)+cot left(frac{3 | vedclass

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(C) We know that $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{\sin^2 	heta + \cos^2 	heta}

Vedclass · Accepted Answer

$2 \sqrt{2-\sqrt{2}}$

Vedclass · Answer

(C) We know that $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin(2 	heta)}$.Here,$	heta = \frac{3 \pi}{16}$,so $2 	heta = \frac{3 \pi}{8}$.Thus,the expression becomes $\frac{2}{\sin(3 \pi / 8)}$.Since $\sin(3 \pi / 8) = \cos(\pi / 2 - 3 \pi / 8) = \cos(\pi / 8)$.Using the half-angle formula,$\cos(\pi / 8) = \sqrt{\frac{1 + \cos(\pi / 4)}{2}} = \sqrt{\frac{1 + 1/\sqrt{2}}{2}} = \sqrt{\frac{\sqrt{2} + 1}{2 \sqrt{2}}} = \frac{\sqrt{2+\sqrt{2}}}{2}$.Therefore,$\frac{2}{\sin(3 \pi / 8)} = \frac{2}{\sqrt{2+\sqrt{2}}/2} = \frac{4}{\sqrt{2+\sqrt{2}}}$.Rationalizing the denominator: $\frac{4}{\sqrt{2+\sqrt{2}}} 	imes \frac{\sqrt{2-\sqrt{2}}}{\sqrt{2-\sqrt{2}}} = \frac{4 \sqrt{2-\sqrt{2}}}{\sqrt{4-2}} = \frac{4 \sqrt{2-\sqrt{2}}}{\sqrt{2}} = 2 \sqrt{2} \sqrt{2-\sqrt{2}} = 2 \sqrt{4-2 \sqrt{2}}$.

$\tan \left(\frac{3 \pi}{16}\right)+\cot \left(\frac{3 \pi}{16}\right)=$

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