Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Prove that Cos 8 ∘ − Sin 8 ∘ Cos 8 ∘ + Sin 8 ∘ = Tan 37 ∘ - Mathematics

Prove that

$\frac{\cos 8^\circ - \sin 8^\circ}{\cos 8^\circ + \sin 8^\circ} = \tan 37^\circ$

#### Solution

$\text{ LHS }= \frac{\cos8^\circ - \sin8^\circ}{\cos8^\circ + \sin8^\circ}$
$= \frac{\frac{\cos8^\circ}{\cos8^\circ} - \frac{\sin8^\circ}{\cos8^\circ}}{\frac{\cos8}{\cos8} + \frac{\sin8}{\cos8}} \left( \text{ Dividing numeraor and denominator by }\cos 8^\circ \right)$
$= \frac{1 - \tan8^\circ}{1 + \tan8^\circ}$
$= \frac{1 - \tan8^\circ}{1 + 1 \times \tan8^\circ}$
$= \frac{\tan45^\circ - \tan8^\circ}{1 + \tan45^\circ \tan8^\circ} \left( \text{ As }\tan 45^\circ = 1 \right)$
$= \tan\left( 45^\circ - 8^\circ \right) \left[\text{ As }\frac{\tan A - \tan B}{1 + \tan A \tan B} = \tan\left( A + B \right) \right]$
$= \tan37^\circ$
= RHS
Hence proved.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 11.3 | Page 19