Question

Prove that :

\[ \frac{\sin(\pi+x)\cos\left(\frac{\pi}{2}+x\right)\tan\left(\frac{3\pi}{2}-x\right)\cot(2\pi-x)} {\sin(2\pi-x)\cos(2\pi+x)\cosec(-x)\sin\left(\frac{3\pi}{2}-x\right)} =1 \]

Solution

\[ \begin{aligned} &\frac{\sin(\pi+x)\cos\left(\frac{\pi}{2}+x\right)\tan\left(\frac{3\pi}{2}-x\right)\cot(2\pi-x)} {\sin(2\pi-x)\cos(2\pi+x)\cosec(-x)\sin\left(\frac{3\pi}{2}-x\right)} \\[8pt] =& \frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)} {(-\sin x)(\cos x)(-\cosec x)(-\cos x)} \\[8pt] =& \frac{-\sin^2x\cot^2x} {-\sin x\cos x\cosec x\cos x} \\[8pt] =& \frac{\sin^2x\cdot\frac{\cos^2x}{\sin^2x}} {\sin x\cos^2x\cdot\frac1{\sin x}} \\[8pt] =& \frac{\cos^2x}{\cos^2x} \\[8pt] =& 1 \end{aligned} \]

Hence Proved.