જો theta = frac{pi}{6} અને x = log left[ cot | vedclass

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(C) આપેલ છે કે $	heta = \frac{\pi}{6}$,તેથી $x = \log \left[ \cot \left( \frac{\pi}{4} + \frac{\pi}{6} ight) ight]$.$x = \log \left[ \cot \left(

Vedclass · Accepted Answer

$-\sqrt{3}$

Vedclass · Answer

(C) આપેલ છે કે $	heta = \frac{\pi}{6}$,તેથી $x = \log \left[ \cot \left( \frac{\pi}{4} + \frac{\pi}{6} ight) ight]$.$x = \log \left[ \cot \left( \frac{5\pi}{12} ight) ight]$.સૂત્ર $\cot(A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$ નો ઉપયોગ કરતા,$\cot \left( \frac{\pi}{4} + \frac{\pi}{6} ight) = \frac{\cot(\pi/4) \cot(\pi/6) - 1}{\cot(\pi/6) + \cot(\pi/4)} = \frac{1 \cdot \sqrt{3} - 1}{\sqrt{3} + 1} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$.આમ,$e^x = \frac{\sqrt{3}-1}{\sqrt{3}+1}$ અને $e^{-x} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$.વ્યાખ્યા $\sinh(x) = \frac{e^x - e^{-x}}{2}$ નો ઉપયોગ કરતા,$\sinh(x) = \frac{1}{2} \left( \frac{\sqrt{3}-1}{\sqrt{3}+1} - \frac{\sqrt{3}+1}{\sqrt{3}-1} ight)$.$\sinh(x) = \frac{1}{2} \left( \frac{(\sqrt{3}-1)^2 - (\sqrt{3}+1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} ight) = \frac{1}{2} \left( \frac{(3+1-2\sqrt{3}) - (3+1+2\sqrt{3})}{3-1} ight) = \frac{1}{2} \left( \frac{-4\sqrt{3}}{2} ight) = -\sqrt{3}$.તેથી,સાચો વિકલ્પ $C$ છે.

જો $\theta = \frac{\pi}{6}$ અને $x = \log \left[ \cot \left( \frac{\pi}{4} + \theta \right) \right]$ હોય,તો $\sinh(x) =$

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