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

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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}} = $

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