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Prove that Cos 11 ∘ + Sin 11 ∘ Cos 11 ∘ − Sin 11 ∘ = Tan 56 ∘

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Question

Prove that

\[\frac{\cos 11^\circ + \sin 11^\circ}{\cos 11^\circ - \sin 11^\circ} = \tan 56^\circ\]
Answer in Brief
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Solution

\[\text{ LHS }= \frac{\cos11^\circ + \sin11^\circ}{\cos11^\circ - \sin11^\circ}\]
\[ = \frac{\frac{\cos11^\circ}{\cos11^\circ} + \frac{\sin11^\circ}{\cos11^\circ}}{\frac{\cos11^\circ}{\cos11^\circ} - \frac{\sin11^\circ}{\cos11^\circ}} \left( \text{ Dividing numerator and denominator by }\cos11^\circ \right)\]
\[ = \frac{1 + \tan11^\circ}{1 - \tan11^\circ}\]
\[ = \frac{1 + \tan11^\circ}{1 - 1 \times \tan11^\circ}\]
\[ = \frac{\tan45^\circ + \tan11^\circ}{1 - \tan45^\circ \tan11^\circ} \left(\text{ As }\tan45^\circ = 1 \right)\]
\[ = \tan\left( 45^\circ + 11^\circ \right) \left[\text{ As }\frac{\tan A + \tan B}{1 - \tan A \tan B} = \tan\left( A + B \right) \right]\]
\[ = \tan56^\circ\]
 = RHS
Hence proved .

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Chapter 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [Page 19]

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RD Sharma Mathematics [English] Class 11
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 11.1 | Page 19

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