Prove the $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$
$L.H.S.$ $=\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$
$=\sin x \cos x[\tan x+\cot x]$
$=\sin x \cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)$
$=(\sin x \cos x)\left[\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos x}\right]$
$=1= R. H. S.$
Which of the following relations is correct
If $A + C = B,$ then $\tan A\,\tan B\,\tan C = $
If $\sin \theta = \frac{{24}}{{25}}$ and $\theta $ lies in the second quadrant, then $\sec \theta + \tan \theta = $
If $\sin x + {\sin ^2}x = 1,$ then ${\cos ^8}x + 2{\cos ^6}x + {\cos ^4}x = $
The equation ${\sec ^2}\theta = \frac{{4xy}}{{{{(x + y)}^2}}}$ is only possible when