$\sum\limits_{r = 1}^{100} {\frac{{\tan \,{2^{r - 1}}}}{{\cos \,{2^r}}}} $ is equal to
$tan\,2^{99} -tan\,1$
$tan\,2^{100}$
$tan\,2^{100} -tan\,1$
none of these
The general solution of $\tan 3x = 1$ is
Let,$S=\left\{\theta \in[0,2 \pi]: 8^{2 \sin ^{2} \theta}+8^{2 \cos ^{2} \theta}=16\right\}$. Then $n ( S )+\sum_{\theta \in S}\left(\sec \left(\frac{\pi}{4}+2 \theta\right) \operatorname{cosec}\left(\frac{\pi}{4}+2 \theta\right)\right)$ is equal to.
If $\sin {\rm{ }}\left( {\frac{\pi }{4}\cot \theta } \right) = \cos {\rm{ }}\left( {\frac{\pi }{4}\tan \theta } \right)\,\,,$ then $\theta = $
The expression $(1 + \tan x + {\tan ^2}x)$ $(1 - \cot x + {\cot ^2}x)$ has the positive values for $x$, given by
If $\cos \theta + \cos 7\theta + \cos 3\theta + \cos 5\theta = 0$, then $\theta $