Question
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
$ = \frac{{\sqrt 3 \cos 20^\circ - \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }} $
$= \frac{{2\left[ {\frac{{\sqrt 3 }}{2}\cos 20^\circ - \frac{1}{2}\sin \,20^\circ } \right]}}{{\frac{2}{2}\sin 20^\circ \cos 20^\circ }}$
$ = \frac{{4\cos (20^\circ + 30^\circ )}}{{\sin 40^\circ }} $
$= \frac{{4\cos 50^\circ }}{{\sin 40^\circ }} = \frac{{4\sin 40^\circ }}{{\sin 40^\circ }} = 4$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\frac{\sum_{k=1}^{12}\left|\alpha_{k+1}-\alpha_k\right|}{\sum_{k=1}^3\left|\alpha_{4 k-1}-\alpha_{4 k-2}\right|}$ का मन है