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Prove That: Tan 225° Cot 405° + Tan 765° Cot 675° = 0

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Question

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

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Solution

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.

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Chapter 5: Trigonometric Functions - Exercise 5.3 [Page 39]

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R.D. Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.3 | Q 2.1 | Page 39

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