Question

Prove the following:

$cos(\frac{3\pi }{2}+x)cos(2\pi +x)$$[cot(\frac{3\pi }{2}+x)+cot(2\pi +x\left)\right]$
= 1

Answer 1

We have

L.H.S. =
cos $(\frac{3\pi }{2}+x)cos(2\pi +x)$$[\text{cot}(\frac{3\pi }{2}+x)+\text{cot}(2\pi +x\left)\right]$

= sin x . cos x [tan x + cot x]

$\left[\begin{array}{c}\because \text{cos}(\frac{3\pi }{2}+x)=\text{sin}x,\text{cos}(2\pi +x)=\text{cos}x\\ cot(\frac{3\pi }{2}-x)=\text{tan}x,\text{cot}(2\pi +x)=\text{cot}x\end{array}\right]$

= $\text{sin x}\text{cos x}\left[\begin{array}{c}\frac{\text{sin x}}{\text{cos x}}+\frac{\text{cos x}}{\text{sin x}}\end{array}\right]$

= $\text{sin x}\text{cos x}\left[\frac{{\text{sin}}^{2}x+{\text{cos}}^{2}x}{\text{sin x cos x}}\right]$

= $\text{sin x}\text{cos x}\left[\frac{1}{\text{sin x cos x}}\right]$

= 1 = R.H.S.