Question

The value of $cot70°+4\cos 70°$ is

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

Answer 1

The correct option is B

$\sqrt{3}$

Explanation for the correct option:

Find the value of the given trigonometric function

Given, $cot70°+4\cos 70°$

$$\begin{gathered} =cot(90°-20°)+4\cos (90°-20°) \\ =\tan {20}^{\circ }+4\sin {20}^{\circ }[\because cot({90}^{\circ }-\theta )=\tan \theta \&\cos ({90}^{\circ }-\theta )=\sin \theta ] \\ =\frac{\sin {20}^{\circ }+4\sin {20}^{\circ }\cos {20}^{\circ }}{\cos {20}^{\circ }} \\ =\frac{\sin {20}^{\circ }+2\sin 40^{\circ }}{\cos {20}^{\circ }} \\ =\frac{\sin {20}^{\circ }+\sin {40}^{\circ }+\sin {40}^{\circ }}{\cos {20}^{\circ }} \\ =\frac{2\sin {30}^{\circ }\cos {10}^{\circ }+\cos ({90}^{\circ }-{40}^{\circ })}{\cos {20}^{\circ }}\left[\because \sin C+\sin D=2\sin \frac{C+D}{2}\cos \frac{C-D}{2}\&\sin \theta =\cos ({90}^{\circ }-\theta )\right] \\ =\frac{2\times {\displaystyle \frac{1}{2}}\cos {10}^{\circ }+\cos {50}^{\circ }}{\cos {20}^{\circ }} \\ =\frac{\cos {10}^{\circ }+\cos {50}^{\circ }}{\cos {20}^{\circ }} \\ =\frac{2\cos {30}^{\circ }\cos {20}^{\circ }}{\cos {20}^{\circ }}\left[\because \cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}\right] \\ =2\times \frac{\sqrt{3}}{2} \\ =\sqrt{3} \end{gathered}$$

Hence, option (B) is the correct answer.