- Home
- Standard 11
- Mathematics
3.Trigonometrical Ratios, Functions and Identities
medium
$\cot {70^o} + 4\cos {70^o} = . . .$
A
$\frac{1}{{\sqrt 3 }}$
B
$\sqrt 3 $
C
$2\sqrt 3 $
D
$\frac{1}{2}$
Solution
(b) Now, $\cot {70^o} + 4\cos {70^o} = \frac{{\cos {{70}^o} + 4\sin {{70}^o}\cos {{70}^o}}}{{\sin {{70}^o}}}$
$ = \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 $.
Standard 11
Mathematics
