Question

The value of the expression $\sqrt{3}\mathrm{cosec}\left(20°\right)-\sec \left(20°\right)$ is equal to

• A. $2$
• B. $\frac{2\sin \left(20°\right)}{\sin \left(40°\right)}$
• C. $4$
• D. $\frac{4\sin \left(20°\right)}{\sin \left(40°\right)}$

Answer 1

The correct option is C

$4$

Solve for the given function:

Given trigonometric expression is $\sqrt{3}\mathrm{cosec}\left(20°\right)-\sec \left(20°\right)$

$$\begin{gathered} \therefore \sqrt{3}\mathrm{cosec}20°-\sec 20° \\ =\frac{\sqrt{3}}{\sin \left(20°\right)}-\frac{1}{\cos \left(20°\right)} \\ =\frac{\sqrt{3}\cos \left(20°\right)-\sin \left(20°\right)}{\cos \left(20°\right)\sin \left(20°\right)} \\ =\frac{4\left[{\displaystyle \frac{\sqrt{3}}{2}\cos \left(20°\right)-\frac{1}{2}\sin \left(20°\right)}\right]}{2\cos \left(20°\right)\sin \left(20°\right)} \\ =\frac{4\left[\sin \left(60°\right)\cos \left(20°\right)-\cos \left(60°\right)\sin \left(20°\right)\right]}{\sin \left(40°\right)}\left[\because \sin \left(2A\right)=2\sin \left(A\right)\cos \left(A\right)\right] \\ =\frac{4\left[\sin \left(60°-20°\right)\right]}{\sin \left(40°\right)}\left[\because \sin \left(A-B\right)=\sin \left(A\right)\cos \left(B\right)-\sin \left(B\right)\cos \left(A\right)\right] \\ =\frac{4\left[\sin \left(40°\right)\right]}{\sin \left(40°\right)} \\ =4 \end{gathered}$$

Hence, option (C) is correct.