કિંમત શોધો: operatorname{cosec}^{-1}left[left | vedclass

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Question

(B) આપણે જાણીએ છીએ કે $\frac{1-	an^2 	heta}{1+	an^2 	heta} = \cos 2	heta$. તેથી,$\frac{	an^2 	heta - 1}{	an^2 	heta + 1} = -\cos 2	heta$. ધા

Vedclass · Accepted Answer

$5 \alpha$

Vedclass · Answer

(B) આપણે જાણીએ છીએ કે $\frac{1-	an^2 	heta}{1+	an^2 	heta} = \cos 2	heta$. તેથી,$\frac{	an^2 	heta - 1}{	an^2 	heta + 1} = -\cos 2	heta$. ધારો કે $	heta = \frac{\alpha-\pi}{4}$. તો $2	heta = \frac{\alpha-\pi}{2} = \frac{\alpha}{2} - \frac{\pi}{2}$. તેથી,$-\cos 2	heta = -\cos(\frac{\alpha}{2} - \frac{\pi}{2}) = -\cos(\frac{\pi}{2} - \frac{\alpha}{2}) = -\sin \frac{\alpha}{2}$. પદાવલિ નીચે મુજબ બને છે: $\operatorname{cosec}^{-1}\left[\left(-\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} \cot 5 \alphaight) \sec \frac{11 \alpha}{2}ight]$ $= \operatorname{cosec}^{-1}\left[\left(-\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} \frac{\cos 5 \alpha}{\sin 5 \alpha}ight) \sec \frac{11 \alpha}{2}ight]$ $= \operatorname{cosec}^{-1}\left[\left(\frac{-\sin 5 \alpha \sin \frac{\alpha}{2} + \cos 5 \alpha \cos \frac{\alpha}{2}}{\sin 5 \alpha}ight) \sec \frac{11 \alpha}{2}ight]$ $= \operatorname{cosec}^{-1}\left[\frac{\cos(5 \alpha + \frac{\alpha}{2})}{\sin 5 \alpha} \cdot \sec \frac{11 \alpha}{2}ight]$ $= \operatorname{cosec}^{-1}\left[\frac{\cos \frac{11 \alpha}{2}}{\sin 5 \alpha} \cdot \frac{1}{\cos \frac{11 \alpha}{2}}ight]$ $= \operatorname{cosec}^{-1}(\operatorname{cosec} 5 \alpha) = 5 \alpha$.

કિંમત શોધો: $\operatorname{cosec}^{-1}\left[\left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}+\cos \frac{\alpha}{2} \cdot \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right]$

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