Question

Prove that:  tan 225° cot 405° + tan 765° cot 675° = 0

Answer 1

LHS = \[\tan225^\circ\cot405^\circ + \tan765^\circ\cot675^\circ\]
\[ = \tan \left( 90^\circ \times 2 + 45^\circ \right)\cot \left( 90^\circ \times 4 + 45^\circ \right) + \tan \left( 90^\circ \times 8 + 45^\circ \right) \cot \left( 90^\circ \times 7 + 45^\circ \right)\]
\[ = \tan \left( 45^\circ \right) \cot \left( 45^\circ \right) + \tan \left( 45^\circ \right)\left[ - \tan \left( 45^\circ \right) \right]\]
\[ = 1 \times 1 + 1 \times \left( - 1 \right)\]
\[ = 1 - 1\]
\[ = 0\]
 = RHS
Hence proved.