Question

sin3x/cosx +cos3x/sinx = 2cot2x​

Answer 1

Given

$\dfrac{\sin 3x}{\cos x}+\dfrac{\cos 3x}{\sin x}=2\cot 2x$

To prove

$\dfrac{\sin 3x}{\cos x}+\dfrac{\cos 3x}{\sin x}=2\cot 2x$

Solution

Formula used

$\sin(a + b) = \sin a \cos b + \cos a \sin b$

$\cos ( a + b )=\cos a \cos b - \sin a \sin b$

$\sin^2 x+\cos^2 x=1$

$\cos x\sin x=\dfrac{\sin 2x}{2}$

Solving left hand side

$\dfrac{\sin 3x}{\cos x}+\dfrac{\cos 3x}{\sin x}=\dfrac{\sin(x+2x)}{\cos x}+\dfrac{\cos(x+2x)}{\sin x}\\ =\dfrac{\sin x\cos 2x+\cos x\sin 2x}{\cos x}+\dfrac{\cos x\cos 2x-\sin x\sin 2x}{\sin x}\\ =\dfrac{\sin x\cos 2x}{\cos x}+\sin 2x+\dfrac{\cos x\cos 2x}{\sin x}-\sin2x\\ =\dfrac{\sin^2 x\cos 2x+\cos^2 x\cos 2x}{\cos x\sin x}\\ =\dfrac{\cos 2x(\sin^2 x+\cos^2 x)}{\cos x\sin x}\\ =\dfrac{\cos 2x}{\cos x\sin x}\\ =\dfrac{\cos 2x}{\dfrac{\sin 2x}{2}}\\ =\dfrac{2\cos 2x}{\sin 2x}\\ =2\cot 2x\\\therefore \dfrac{\sin 3x}{\cos x}+\dfrac{\cos 3x}{\sin x}=2\cot 2x$

So, left hand side is equal to right hand side.

Hence, proved.