Question

Prove that :

\[ \tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} = 0 \]

Solution

Reducing the angles,

\[ \tan\frac{5\pi}{4} = \tan\frac{\pi}{4} = 1 \]

\[ \cot\frac{9\pi}{4} = \cot\frac{\pi}{4} = 1 \]

\[ \tan\frac{17\pi}{4} = \tan\frac{\pi}{4} = 1 \]

\[ \cot\frac{15\pi}{4} = \cot\frac{7\pi}{4} = -1 \]

Substituting these values,

\[ \begin{aligned} &\tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} \\[8pt] =& (1)(1)+(1)(-1) \\[8pt] =& 1-1 \\[8pt] =& 0 \end{aligned} \]

Hence Proved.