cot {70^o} + 4cos {70^o} = . . . | vedclass

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Question

(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 {{14

Vedclass · Accepted Answer

$\sqrt 3 $

Vedclass · Answer

(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 $.

$\cot {70^o} + 4\cos {70^o} = . . .$

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