Question

The value of $\cot {70}^{\circ }+4\cos {70}^{\circ }$ is

• A. $\frac{1}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $2\sqrt{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is B $\sqrt{3}$

$a=\cot {70}^{}+4\cos {70}^{}$

$\Rightarrow a=\cot ({90}^{}-{20}^{})+4\cos ({90}^{}-{20}^{})$

$\Rightarrow a=\tan {20}^{}+4\sin {20}^{}$ $[\because \cot (90-\theta )=\tan \theta ,$$\cos (90-\theta )=\sin \theta ]$

$\Rightarrow a=\frac{\sin {20}^{}+4\sin {20}^{}\cos {20}^{}}{\cos {20}^{}}$

$\Rightarrow a=\frac{\sin {20}^{}+2\sin {40}^{}}{\cos {20}^{}}=\frac{\sin {20}^{}+\sin {40}^{}+\sin {40}^{}}{\cos {20}^{}}$

$\Rightarrow a=\frac{2\sin {30}^{}\cos {10}^{}+\cos ({90}^{}-{40}^{})}{\cos {20}^{}}$ $[\because \sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}]$

$\Rightarrow a=\frac{\cos {10}^{}+\cos {50}^{}}{\cos {20}^{}}$

$=\frac{2\cos {30}^{}\cos {20}^{}}{\cos {20}^{}}$ $[\because \cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}]$

$\Rightarrow a=2\times \frac{\sqrt{3}}{2}$

$\Rightarrow a=\sqrt{3}$

Ans: B