Question

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

Answer 1

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

=$sin\,x\,cos\,x[tan\,x+cot\,x]$

$=sin\,x\,cos\,x(\frac{sin\\.x}{cos\,x}+\frac{cos\\.x}{sin\,x})$

$=(sin\,x\,cos\,x)[\frac{sin^2+cos^2\,x}{sin\,x\,cos\,x}]$

$=1=R.H.S.$