The value of frac{tan 70^o - tan 20^o}{tan 50 | vedclass

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(B) We have $\frac{	an 70^o - 	an 20^o}{	an 50^o}$.Using $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we get:$= \frac{\frac{\sin 70^o}{\cos 70^o

Vedclass · Accepted Answer

$2$

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(B) We have $\frac{	an 70^o - 	an 20^o}{	an 50^o}$.Using $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we get:$= \frac{\frac{\sin 70^o}{\cos 70^o} - \frac{\sin 20^o}{\cos 20^o}}{	an 50^o}$$= \frac{\sin 70^o \cos 20^o - \cos 70^o \sin 20^o}{\cos 70^o \cos 20^o 	an 50^o}$Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$,the numerator becomes $\sin(70^o - 20^o) = \sin 50^o$.$= \frac{\sin 50^o}{\cos 70^o \cos 20^o 	an 50^o}$Since $	an 50^o = \frac{\sin 50^o}{\cos 50^o}$,we have:$= \frac{\sin 50^o \cos 50^o}{\cos 70^o \cos 20^o \sin 50^o} = \frac{\cos 50^o}{\cos 70^o \cos 20^o}$Multiply numerator and denominator by $2$:$= \frac{2 \cos 50^o}{2 \cos 70^o \cos 20^o}$Using $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$:$= \frac{2 \cos 50^o}{\cos(70^o + 20^o) + \cos(70^o - 20^o)}$$= \frac{2 \cos 50^o}{\cos 90^o + \cos 50^o}$Since $\cos 90^o = 0$:$= \frac{2 \cos 50^o}{0 + \cos 50^o} = \frac{2 \cos 50^o}{\cos 50^o} = 2$.

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

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