Find the value of $\tan \frac{\pi}{8}$
Let $x=\frac{\pi}{8} .$ Then $2 x=\frac{\pi}{4}$
Now $\tan 2 x=\frac{2 \tan x}{1-\tan ^{2} x}$
or $\tan \frac{\pi}{4}=\frac{2 \tan \frac{\pi}{8}}{1-\tan ^{2} \frac{\pi}{8}}$
Let $y=\tan \frac{\pi}{8} .$ Then $1=\frac{2 y}{1-y^{2}}$
or $y^{2}+2 y-1=0$
Therefore $y=\frac{-2 \pm 2 \sqrt{2}}{2}=-1 \pm \sqrt{2}$
since $\frac{\pi}{8}$ lies in the first quadrant, $y=\tan \frac{\pi}{8}$ is positve. Hence
$\tan \frac{\pi}{8}=\sqrt{2}-1$
Find the general solution of the equation $\cos 4 x=\cos 2 x$
If $m$ and $n$ respectively are the numbers of positive and negative value of $\theta$ in the interval $[-\pi, \pi]$ that satisfy the equation $\cos 2 \theta \cos \frac{\theta}{2}=\cos 3 \theta \cos \frac{9 \theta}{2}$, then $mn$ is equal to $.............$.
The most general value of $\theta $ which will satisfy both the equations $\sin \theta = - \frac{1}{2}$ and $\tan \theta = \frac{1}{{\sqrt 3 }}$ is
Let $\theta, \phi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1, \tan (2 \pi-\theta)>0$ and $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. Then $\phi$ cannot satisfy
$(A)$ $0 < \phi<\frac{\pi}{2}$ $(B)$ $\frac{\pi}{2} < \phi<\frac{4 \pi}{3}$
$(C)$ $\frac{4 \pi}{3} < \phi<\frac{3 \pi}{2}$ $(D)$ $\frac{3 \pi}{2} < \phi < 2 \pi$
The solution set of $(5 + 4\cos \theta )(2\cos \theta + 1) = 0$ in the interval $[0,\,\,2\pi ]$ is