જો x=log _eleft[cot left(frac{pi}{4}+thetarig | vedclass

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(C) આપણી પાસે $x=\log _e \left[\cot \left(\frac{\pi}{4}+	hetaight)ight]$ છે.વિધાન $I$ માટે:$\cosh x = \frac{e^x+e^{-x}}{2} = \frac{\cot \left(\fr

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$I$ અને $II$ બંને સાચા છે

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(C) આપણી પાસે $x=\log _e \left[\cot \left(\frac{\pi}{4}+	hetaight)ight]$ છે.વિધાન $I$ માટે:$\cosh x = \frac{e^x+e^{-x}}{2} = \frac{\cot \left(\frac{\pi}{4}+	hetaight) + 	an \left(\frac{\pi}{4}+	hetaight)}{2}$$= \frac{\frac{\cos(\pi/4+	heta)}{\sin(\pi/4+	heta)} + \frac{\sin(\pi/4+	heta)}{\cos(\pi/4+	heta)}}{2} = \frac{\cos^2(\pi/4+	heta) + \sin^2(\pi/4+	heta)}{2 \sin(\pi/4+	heta) \cos(\pi/4+	heta)}$$= \frac{1}{\sin(2(\pi/4+	heta))} = \frac{1}{\sin(\pi/2+2	heta)} = \frac{1}{\cos 2	heta} = \sec 2	heta$.આમ,વિધાન $I$ સાચું છે.વિધાન $II$ માટે:$\sinh x = \frac{e^x-e^{-x}}{2} = \frac{\cot \left(\frac{\pi}{4}+	hetaight) - 	an \left(\frac{\pi}{4}+	hetaight)}{2}$$= \frac{\frac{\cos^2(\pi/4+	heta) - \sin^2(\pi/4+	heta)}{\sin(\pi/4+	heta) \cos(\pi/4+	heta)}}{2} = \frac{\cos(2(\pi/4+	heta))}{\sin(2(\pi/4+	heta))}$$= \cot(\pi/2+2	heta) = -	an 2	heta$.આમ,વિધાન $II$ પણ સાચું છે.

જો $x=\log _e\left[\cot \left(\frac{\pi}{4}+\theta\right)\right]$ અને $\theta \in\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)$ હોય,તો નીચેના વિધાનો ધ્યાનમાં લો:
$I$. $\cosh x=\sec 2 \theta$
$II$. $\sinh x=-\tan 2 \theta$

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જો $x=\log _e\left[\cot \left(\frac{\pi}{4}+\theta\right)\right]$ અને $\theta \in\left(\frac{-\pi}{4}, \frac{\pi}{4}\right)$ હોય,તો નીચેના વિધાનો ધ્યાનમાં લો:$I$. $\cosh x=\sec 2 \theta$$II$. $\sinh x=-\tan 2 \theta$