STD 11 - 3. trignometrical ratios,functions and identities — JEE कक्षा 11 साइन्स — Question

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

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

$=  \frac{2}{2} \times \frac{{\sin ({{70}^o} - {{20}^o})\cos {{50}^o}}}{{\cos {{70}^o}\cos {{20}^o}\sin {{50}^o}}}$

$=  \frac{{2\sin {{50}^o}\cos {{50}^o}}}{{2\cos {{70}^o}\cos {{20}^o}\sin {{50}^o}}}$

$=  \frac{{2\cos {{50}^o}}}{{\cos {{90}^o} + \cos {{50}^o}}}$

$= \frac{{2\cos {{50}^o}}}{{0 + \cos {{50}^o}}}$

$= 2.$