If $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ and $M=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right),$ then
$M =\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$
$L =\frac{1}{4 \sqrt{2}}-\frac{1}{4} \cos \frac{\pi}{8}$
$M =\frac{1}{4 \sqrt{2}}+\frac{1}{4} \cos \frac{\pi}{8}$
$L =-\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$
If $tanA + cotA = 4$, then $tan^4A + cot^4A$ is equal to
The solution of the equation ${\cos ^2}x - 2\cos x = $ $4\sin x - \sin 2x,$ $\,(0 \le x \le \pi )$ is
If $\cos \,\alpha + \cos \,\beta = \frac{3}{2}$ and $\sin \,\alpha + \sin \,\beta = \frac{1}{2}$ and $\theta $ is the the arithmetic mean of $\alpha $ and $\beta $ , then $\sin \,2\theta + \cos \,2\theta $ is equal to
The angles $\alpha, \beta, \gamma$ of a triangle satisfy the equations $2 \sin \alpha+3 \cos \beta=3 \sqrt{2}$ and $3 \sin \beta+2 \cos \alpha=1$. Then, angle $\gamma$ equals
General solution of $eq^n\, 2tan\theta \, -\, cot\theta =\, -1$ is