3. trignometrical ratios,functions and identities — ગણિત ધોરણ 11 સાયન્સ — Question

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

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

$ = \frac{{\sin ({{20}^o} - {{70}^o})}}{{\frac{1}{2}[\cos ({{70}^o} + {{20}^o}) + \cos ({{70}^o} - {{20}^o})]}}$$ + 2\tan {50^o}$

$ = \frac{{2\sin ( - {{50}^o})}}{{\cos {{90}^o} + \cos {{50}^o}}} + 2\tan {50^o}$

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

$ = - 2\tan {50^o} + 2\tan {50^o} = 0$.