The value of frac{tan 70^circ - tan 20^circ}{ | vedclass

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Question

(B) We are given the expression $\frac{	an 70^\circ - 	an 20^\circ}{	an 50^\circ}$.Using the identity $	an 	heta = \frac{\sin 	heta}{\cos 	heta

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We are given the expression $\frac{	an 70^\circ - 	an 20^\circ}{	an 50^\circ}$.Using the identity $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we have:$\frac{\frac{\sin 70^\circ}{\cos 70^\circ} - \frac{\sin 20^\circ}{\cos 20^\circ}}{\frac{\sin 50^\circ}{\cos 50^\circ}}$Simplifying the numerator:$= \frac{\frac{\sin 70^\circ \cos 20^\circ - \cos 70^\circ \sin 20^\circ}{\cos 70^\circ \cos 20^\circ}}{\frac{\sin 50^\circ}{\cos 50^\circ}}$Using the sine subtraction formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$= \frac{\sin(70^\circ - 20^\circ)}{\cos 70^\circ \cos 20^\circ} 	imes \frac{\cos 50^\circ}{\sin 50^\circ}$$= \frac{\sin 50^\circ}{\cos 70^\circ \cos 20^\circ} 	imes \frac{\cos 50^\circ}{\sin 50^\circ} = \frac{\cos 50^\circ}{\cos 70^\circ \cos 20^\circ}$Multiply numerator and denominator by $2$:$= \frac{2 \cos 50^\circ}{2 \cos 70^\circ \cos 20^\circ}$Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$:$= \frac{2 \cos 50^\circ}{\cos(70^\circ + 20^\circ) + \cos(70^\circ - 20^\circ)}$$= \frac{2 \cos 50^\circ}{\cos 90^\circ + \cos 50^\circ}$Since $\cos 90^\circ = 0$:$= \frac{2 \cos 50^\circ}{0 + \cos 50^\circ} = 2$.

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

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