Question

Find the value of x: $\cos \left(\frac{3\pi }{2}+x\right)\cos \left(2\pi +x\right)\left[\cot \left(\frac{3\pi }{2}-x\right)+\cot \left(2\pi +x\right)\right]=1$

Answer 1

At first we solve,

$\begin{array}{c}\cos \left(\frac{3\pi }{2}+x\right)putting\pi ={180}^{\circ }\\ =\cos \left(\frac{3\times {180}^{\circ }}{2}+x\right)\\ =\cos \left({270}^{\circ }+x\right)\\ =\cos \left({360}^{\circ }-{90}^{\circ }+x\right)\\ =\cos \left(2\pi +\left(x-{90}^{\circ }\right)\right)\\ =\cos \left(x-{90}^{\circ }\right)\left[\cos \left(2\pi +x\right)=\cos x\right]\\ =\cos \left(-\left({90}^{\circ }-x\right)\right)\\ =\cos \left({90}^{\circ }-x\right)\left(\cos \left(-x\right)=\cos x\right)\\ =\sin x\\ Now,\cos \left(2\pi +x\right)=\cos x\\ \&\cot \left(2\pi +x\right)=\cot x\\ Now,solve\\ \cot \left(\frac{3\pi }{2}-x\right)\\ =\cot \left(\frac{3\times {180}^{\circ }}{2}-x\right)\left(put\pi ={180}^{\circ }\right)\\ =\cot \left({270}^{\circ }-x\right)\\ =\cot \left({360}^{\circ }-{90}^{\circ }-x\right)\\ =\cot \left(2\pi -{90}^{\circ }-x\right)\\ =\cot \left(2\pi \left(x+{90}^{\circ }\right)\right)\\ =-\cot \left(x+90\right)\left[\cot 2\pi -x=-\cot x\right]\\ =-\left(-\tan x\right)\left[\cot \left(90+\theta \right)=-\tan \theta \right]\\ =\tan x\\ \\ Nowputtingvaluesinequation\\ L.H.S=\cos \left(\frac{3\pi }{2}+x\right)\cos \left(2\pi +x\right)\left(\cot \left(\frac{3\pi }{2}-x\right)+\cot \left(2\pi +x\right)\right)\\ =\left(\sin x\right)\times \left(\cos x\right)\times \left[\tan x+\cot x\right]\\ =\left(\sin x\cos x\right)\times \left[\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right]\\ =\left(\sin x\cos x\right)\times \frac{{\cos }^{2}x+{\sin }^{2}x}{\left(\sin x\cos x\right)}\\ ={\cos }^{2}x+{\sin }^{2}x\\ =1\\ =R.H.S\\ Hence,proved.\end{array}$