माना कि frac{pi}{2}

Question

$x \in\left(\frac{\pi}{2}, \pi\right)$

$\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$

Vedclass · Accepted Answer

$\frac{\sqrt{11}+1}{2 \sqrt{3}}$

Vedclass · Answer

$x \in\left(\frac{\pi}{2}, \piight)$

$\left(\sin \frac{11 x}{2}ight)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}ight)(\sin 6 x+\cos 6 x)$

$=\left\{\sin 6 x \sin \frac{11 x}{2}+\cos \frac{11 x}{2} \cos 6 xight\}$

$=\cos \left(6 x-\frac{11 x}{2}ight)+\sin \left(6 x-\frac{11 x}{2}ight)$

$=\cos \frac{x}{2}+\sin \frac{x}{2}$

$=\frac{1}{2 \sqrt{3}}+\frac{\sqrt{11}}{2 \sqrt{3}}$

$=\frac{\sqrt{11}+1}{2 \sqrt{3}} \Rightarrow 	ext { Option (B) is correct. }$

$\cot x=-\frac{5}{\sqrt{11}}$

$\frac{1-	an ^2 \frac{x}{2}}{2 	an \frac{x}{2}}=-\frac{5}{\sqrt{11}}$

$	an \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}}$

$	an \frac{x}{2}=\sqrt{11}$, As $\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

माना कि $\frac{\pi}{2} < x < \pi$ इस प्रकार है कि $\cot x=\frac{-5}{\sqrt{11}}$ है। तब

$\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$ बराबर है

Similar Questions

यदि ${\rm{cosec}}\theta = \frac{{p + q}}{{p - q}},$ तब $\cot \,\left( {\frac{\pi }{4} + \frac{\theta }{2}} \right) = $

यदि $\tan \beta = \cos \theta \tan \alpha ,$ तब ${\tan ^2}\frac{\theta }{2} = $

$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $

$\sin 12^\circ \sin 48^\circ \sin 54^\circ = $

$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} =$

माना कि $\frac{\pi}{2} < x < \pi$ इस प्रकार है कि $\cot x=\frac{-5}{\sqrt{11}}$ है। तब $\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$ बराबर है