UPSEE Solved Model Paper 6

We can solve this problem using the trigonometric function of Sum and Difference Compound Angles: An algebraic sum of two or more angles is called a compound angle tan (A - B) = tan A - tan B/1 + tanA tanB By using this formula we can find the answer. We have $\frac{\sin 79^{\circ} - \sin 11^{\circ}}{\sin 79^{\circ} + \sin 11^{\circ}}$ $\frac{\sin 79^{\circ} - \sin 11^{\circ}}{\sin 79^{\circ} + \sin 11^{\circ}}$ = $\frac{\cos 11^{\circ} - \sin 11^{\circ}}{\cos 11^{\circ} + \sin 11^{\circ}}$ (Since , sin (90 - θ) = cos θ ; sin 79° = sin (90 - 11) = cos 11°) Divide the numerator and denominator by cos 11° = $\frac{ \frac{\cos 11^{\circ} - \sin 11^{\circ}}{\cos 11^{\circ}} }{ \frac{\cos 11^{\circ} + \sin 11^{\circ}}{\cos 11^{\circ}} }$ = $\frac{1 - \frac{\sin 11^{\circ}}{\cos 11^{\circ}}}{1 + \frac{\sin 11^{\circ}}{\cos 11^{\circ}}}$ = $\frac{1 - \tan 11^{\circ}}{1 + \tan 11^{\circ}}$ = $\frac{\tan 45^{\circ} - \tan 11^{\circ}}{1 + \tan 45^{\circ} \times \tan 11^{\circ}}$ (since, tan 45 = 1) = tan (45° - 11°) Since $\frac{\tan A - \tan B}{1 + \tan A \tan B}$ = tan (A - B) = tan 34° ∴ $\frac{\sin 79^{\circ} - \sin 11^{\circ}}{\sin 79^{\circ} + \sin 11^{\circ}}$ = tan 34°