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

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(C) આપેલ છે,$x=\log _e\left[\cot \left(\frac{\pi}{4}+	hetaight)ight]$$\Rightarrow e^x =\cot \left(\frac{\pi}{4}+	hetaight) = \frac{\cos 	heta

Vedclass · Accepted Answer

$I$ અને $II$ બંને સાચા છે

Vedclass · Answer

(C) આપેલ છે,$x=\log _e\left[\cot \left(\frac{\pi}{4}+	hetaight)ight]$$\Rightarrow e^x =\cot \left(\frac{\pi}{4}+	hetaight) = \frac{\cos 	heta-\sin 	heta}{\cos 	heta+\sin 	heta}$હવે,$\cosh x = \frac{e^x+e^{-x}}{2} = \frac{1}{2}\left[\frac{\cos 	heta-\sin 	heta}{\cos 	heta+\sin 	heta} + \frac{\cos 	heta+\sin 	heta}{\cos 	heta-\sin 	heta}ight]$$= \frac{1}{2}\left[\frac{(\cos 	heta-\sin 	heta)^2+(\cos 	heta+\sin 	heta)^2}{\cos^2 	heta-\sin^2 	heta}ight] = \frac{1}{2}\left[\frac{2(\cos^2 	heta+\sin^2 	heta)}{\cos 2 	heta}ight] = \frac{1}{\cos 2 	heta} = \sec 2 	heta$તેથી,વિધાન $I$ સાચું છે.હવે,$\sinh x = \frac{e^x-e^{-x}}{2} = \frac{1}{2}\left[\frac{\cos 	heta-\sin 	heta}{\cos 	heta+\sin 	heta} - \frac{\cos 	heta+\sin 	heta}{\cos 	heta-\sin 	heta}ight]$$= \frac{1}{2}\left[\frac{(\cos 	heta-\sin 	heta)^2-(\cos 	heta+\sin 	heta)^2}{\cos^2 	heta-\sin^2 	heta}ight] = \frac{1}{2}\left[\frac{-4 \sin 	heta \cos 	heta}{\cos 2 	heta}ight] = \frac{-\sin 2 	heta}{\cos 2 	heta} = -	an 2 	heta$તેથી,વિધાન $II$ સાચું છે.

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