Question

Prove that:

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

 

Answer 1

\[ \frac{5\pi}{4} = 225^\circ, \frac{9\pi}{4} = 405^\circ, \frac{17\pi}{4} = 765^\circ, \frac{15\pi}{4} = 675^\circ\]
LHS = \[\tan 225^\circ\cot 405^\circ + \tan 765^\circ \cot 675^\circ\]
\[ = \tan\left( 90^\circ \times 2 + 45^\circ \right)\cot\left( 90^\circ \times 4 + 45^\circ \right) + \tan\left( 90^\circ \times 8 + 45^\circ \right) \cot\left( 90^\circ \times 7 + 45^\circ \right) \]
\[ = \tan 45^\circ\cot 45^\circ + \tan 45^\circ \left[ - \tan45^\circ \right]\]
\[ = \tan 45^\circ\cot 45^\circ - \tan 45^\circ \tan 45^\circ\]
\[ = 1 \times 1 - 1 \times 1\]
\[ = 1 - 1\]
\[ = 0\]
 = RHS
Hence proved.