MCQ
$\cot {70^o} + 4\cos {70^o} = . . .$
- A$\frac{1}{{\sqrt 3 }}$
- ✓$\sqrt 3 $
- C$2\sqrt 3 $
- D$\frac{1}{2}$
$ = \frac{{\cos {{70}^o} + 2\sin {{140}^o}}}{{\sin {{70}^o}}} $
$= \frac{{\cos {{70}^o} + 2\sin ({{180}^o} - {{40}^o})}}{{\sin {{70}^o}}}$
$ = \frac{{\sin {{20}^o} + \sin {{40}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}} $
$= \frac{{2\sin {{30}^o}\cos {{10}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}}$
$ = \frac{{\sin {{80}^o} + \sin {{40}^o}}}{{\sin {{70}^o}}} $
$= \frac{{2\sin {{60}^o}\cos {{20}^o}}}{{\sin {{70}^o}}} = \sqrt 3 $.
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