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

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(C) हम जानते हैं कि $	an 	heta + \cot 	heta = \frac{2}{\sin(2 	heta)}$.यहाँ,$	heta = \frac{3 \pi}{16}$,इसलिए $2 	heta = \frac{3 \pi}{8}$.अतः,व्य

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

Vedclass · Answer

(C) हम जानते हैं कि $	an 	heta + \cot 	heta = \frac{2}{\sin(2 	heta)}$.यहाँ,$	heta = \frac{3 \pi}{16}$,इसलिए $2 	heta = \frac{3 \pi}{8}$.अतः,व्यंजक $\frac{2}{\sin(3 \pi / 8)}$ हो जाता है।$\sin(3 \pi / 8) = \cos(\pi / 8) = \frac{\sqrt{2+\sqrt{2}}}{2}$.इसलिए,$\frac{2}{\sin(3 \pi / 8)} = \frac{4}{\sqrt{2+\sqrt{2}}} = 2 \sqrt{2} \sqrt{2-\sqrt{2}}$.

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

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