$6$ रेडियन को डिग्री माप में बदलिए।
We know that $\pi$ radian $=180^{\circ}$
Hence $6 \text { radians }=\frac{180}{\pi} \times 6 \text { degree }=\frac{1080 \times 7}{22} \text { degree }$
${ = 343\frac{7}{{11}}{\text{ degree }} = {{343}^\circ } + \frac{{7 \times 60}}{{11}}{\text{ minute }}\left[ {{\text{ as }}{1^\circ } = {{60}^\prime }} \right]}$
${ = {{343}^\circ } + {{38}^\prime } + \frac{2}{{11}}{\text{ minute }}}$ ${[{\text{as }}{{\text{1}}^\prime }{\text{ = 6}}{{\text{0}}^{\prime \prime }}]}$
${ = {{343}^\circ } + {{38}^\prime } + {{10.9}^{\prime \prime }}}$ $=343^{\circ} 38^{\prime} 11^{\prime \prime}$ approximately
Hence $6$ radians $=343^{\circ} 38^{\prime} 11^{\prime \prime}$ approximately.
यदि $p = \frac{{2\sin \,\theta }}{{1 + \cos \theta + \sin \theta }}$,तथा $q = \frac{{\cos \theta }}{{1 + \sin \theta }},$ तो
यदि $\tan \theta = \frac{{20}}{{21}},$ cos$\theta$
यदि $\frac{{3\pi }}{4} < \alpha < \pi ,$ हो, तब $\sqrt {{\rm{cose}}{{\rm{c}}^2}\alpha + 2\cot \alpha } $ बराबर है
निम्न में कौन सा सही है
यदि $x\sin 45^\circ {\cos ^2}60^\circ = \frac{{{{\tan }^2}60^\circ {\rm{cosec}}30^\circ }}{{\sec 45^\circ {{\cot }^2}30^\circ }},$ तब $x = $