$\frac{{\sin \theta }}{{1 - \cot \theta }} + \frac{{\cos \theta }}{{1 - \tan \theta }} = $
$0$
$1$
$\cos \theta - \sin \theta $
$\cos \theta + \sin \theta $
If $(\sec \alpha + \tan \alpha )(\sec \beta + \tan \beta )(\sec \gamma + \tan \gamma )$$ = \tan \alpha \tan \beta \tan \gamma $, then $(\sec \alpha - \tan \alpha )(\sec \beta - \tan \beta )$$(\sec \gamma - \tan \gamma ) = $
If $a\,{\cos ^3}\alpha + 3a\,\cos \alpha \,{\sin ^2}\alpha = m$ and $a\,{\sin ^3}\alpha + 3a\,{\cos ^2}\alpha \sin \alpha = n,$ then ${(m + n)^{2/3}} + {(m - n)^{2/3}}$ is equal to
Find the value of the trigonometric function $\tan \frac{19 \pi}{3}$.
Prove that :
$\cot ^{2} \frac{\pi}{6}+\cos ec \,\frac{5 \pi}{6}+3 \tan ^{2}\, \frac{\pi}{6}=6$
The equation ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ is only possible when