frac{tan 52^{circ} - tan 38^{circ}}{tan 14^{c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that $	an A - 	an B = \frac{\sin(A - B)}{\cos A \cos B}$.Applying this to the numerator: $	an 52^{\circ} - 	an 38^{\circ} = \frac{\sin

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We know that $	an A - 	an B = \frac{\sin(A - B)}{\cos A \cos B}$.Applying this to the numerator: $	an 52^{\circ} - 	an 38^{\circ} = \frac{\sin(52^{\circ} - 38^{\circ})}{\cos 52^{\circ} \cos 38^{\circ}} = \frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ}}$.Now,the expression becomes $\frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ} 	an 14^{\circ}} = \frac{\sin 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ} \frac{\sin 14^{\circ}}{\cos 14^{\circ}}} = \frac{\cos 14^{\circ}}{\cos 52^{\circ} \cos 38^{\circ}}$.Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$,we have $2 \cos 52^{\circ} \cos 38^{\circ} = \cos(52^{\circ} + 38^{\circ}) + \cos(52^{\circ} - 38^{\circ}) = \cos 90^{\circ} + \cos 14^{\circ} = 0 + \cos 14^{\circ} = \cos 14^{\circ}$.Thus,the expression is $\frac{\cos 14^{\circ}}{\frac{1}{2} \cos 14^{\circ}} = 2$.

$\frac{\tan 52^{\circ} - \tan 38^{\circ}}{\tan 14^{\circ}} = $

Similar Questions

The value of $\frac{\tan 70^o - \tan 20^o}{\tan 50^o} = $

If $\cos \theta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5},$ where $\theta$ and $\phi$ are positive acute angles,then $\cos \frac{\theta - \phi}{2} = $

The value of $\sin 47^\circ + \sin 61^\circ - \sin 11^\circ - \sin 25^\circ = $

$1 + \cos 2x + \cos 4x + \cos 6x = $

If $\sin x + \sin y = \frac{1}{2}$ and $\cos x + \cos y = 1$,then $\tan(x + y) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module