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

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Question

(b) $\frac{{	an {{70}^o} - 	an {{20}^o}}}{{	an {{50}^o}}}$

$=  \frac{{\frac{{\sin {{70}^o}}}{{\cos {{70}^o}}} - \frac{{\sin {{20}^o}}}{{\co

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(b) $\frac{{	an {{70}^o} - 	an {{20}^o}}}{{	an {{50}^o}}}$

$=  \frac{{\frac{{\sin {{70}^o}}}{{\cos {{70}^o}}} - \frac{{\sin {{20}^o}}}{{\cos {{20}^o}}}}}{{\frac{{\sin {{50}^o}}}{{\cos {{50}^o}}}}}$

$=  \frac{{\frac{{\sin {{70}^o}\cos {{20}^o} - \cos {{70}^o}\sin {{20}^o}}}{{\cos {{70}^o}\cos {{20}^o}}}}}{{\frac{{\sin {{50}^o}}}{{\cos {{50}^o}}}}}$

$=  \frac{2}{2} 	imes \frac{{\sin ({{70}^o} - {{20}^o})\cos {{50}^o}}}{{\cos {{70}^o}\cos {{20}^o}\sin {{50}^o}}}$

$=  \frac{{2\sin {{50}^o}\cos {{50}^o}}}{{2\cos {{70}^o}\cos {{20}^o}\sin {{50}^o}}}$

$=  \frac{{2\cos {{50}^o}}}{{\cos {{90}^o} + \cos {{50}^o}}}$

$= \frac{{2\cos {{50}^o}}}{{0 + \cos {{50}^o}}}$

$= 2.$

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

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