Question

The value of $\sqrt{3}\csc {{20}^{0}}-\sec {{20}^{0}}$ is.
(a) 4.5
(b) 4
(c) 8
(d) 9

Answer 1

Hint: For solving this question we will use some trigonometric formulas like \[\cos \left( A+B \right)=\cos A\cdot \cos B-\sin A\cdot \sin B\] , \[\sin 2\theta =2\sin \theta \cos \theta \] , \[\cos \theta =\sin \left( 90-\theta \right)\] and some common trigonometric ratios like \[\cos {{30}^{0}}=\dfrac{\sqrt{3}}{2}\] and \[\sin {{30}^{0}}=\dfrac{1}{2}\] . After that, we will simplify the given term then put the proper values and solve for the correct answer.

Complete step-by-step answer:
Given:
We have to find the value of $\sqrt{3}\csc {{20}^{0}}-\sec {{20}^{0}}$ .
Now, here either we put directly the value of $\csc {{20}^{0}}$ , $\sec {{20}^{0}}$ and find the value of the given term. Rather than putting value, we will first simplify the given term. We will use the following formulae in solving this question:
\[\begin{align}
  & \csc \theta =\dfrac{1}{\sin \theta }.............................................................\left( 1 \right) \\
 & \sec \theta =\dfrac{1}{\cos \theta }.............................................................\left( 2 \right) \\
 & \cos {{30}^{0}}=\dfrac{\sqrt{3}}{2}.............................................................\left( 3 \right) \\
 & \sin {{30}^{0}}=\dfrac{1}{2}.................................................................\left( 4 \right) \\
 & \cos \left( A+B \right)=\cos A\cdot \cos B-\sin A\cdot \sin B......................\left( 5 \right) \\
 & \sin 2\theta =2\sin \theta \cos \theta ....................................................\left( 6 \right) \\
 & \cos \theta =\sin \left( 90-\theta \right)......................................................\left( 7 \right) \\
\end{align}\]
Now, simplifying the given term. Then,
$\sqrt{3}\csc {{20}^{0}}-\sec {{20}^{0}}$
Now, use equation (1) and equation (2) in the above term. Then,
$\begin{align}
  & \sqrt{3}\csc {{20}^{0}}-\sec {{20}^{0}} \\
 & \Rightarrow \dfrac{\sqrt{3}}{\sin {{20}^{0}}}-\dfrac{1}{\cos {{20}^{0}}} \\
 & \Rightarrow \dfrac{\sqrt{3}\cos {{20}^{0}}-\sin {{20}^{0}}}{\sin {{20}^{0}}\cos {{20}^{0}}} \\
 & \Rightarrow \dfrac{2\left( \dfrac{\sqrt{3}}{2}\cos {{20}^{0}}-\dfrac{1}{2}\sin {{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\
\end{align}$
Now, substituting \[\dfrac{\sqrt{3}}{2}=\cos {{30}^{0}}\] from equation (3) and \[\dfrac{1}{2}=\sin {{30}^{0}}\] from equation (4) in the above term. Then,
$\begin{align}
  & \dfrac{2\left( \dfrac{\sqrt{3}}{2}\cos {{20}^{0}}-\dfrac{1}{2}\sin {{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\
 & \Rightarrow \dfrac{2\left( \cos {{30}^{0}}\cos {{20}^{0}}-\sin {{30}^{0}}\sin {{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\
\end{align}$
Now, using the formula from the equation (5) with the value of $A={{30}^{0}}$ and $B={{20}^{0}}$ . Then,
$\begin{align}
  & \dfrac{2\left( \cos {{30}^{0}}\cos {{20}^{0}}-\sin {{30}^{0}}\sin {{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\
 & \Rightarrow \dfrac{2\cos \left( {{30}^{0}}+{{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\
 & \Rightarrow \dfrac{4\cos {{50}^{0}}}{2\sin {{20}^{0}}\cos {{20}^{0}}} \\
\end{align}$
Now, using the formula from the equation (6) with the value of $\theta ={{20}^{0}}$ . Then,
$\begin{align}
  & \dfrac{4\cos {{50}^{0}}}{2\sin {{20}^{0}}\cos {{20}^{0}}} \\
 & \Rightarrow \dfrac{4\cos {{50}^{0}}}{\sin {{40}^{0}}} \\
\end{align}$
Now, substituting $\sin {{40}^{0}}=\cos {{50}^{0}}$ in the above term by using the formula from equation (7) with the value of $\theta ={{50}^{0}}$ . Then,
$\begin{align}
  & \dfrac{4\cos {{50}^{0}}}{\sin {{40}^{0}}} \\
 & \Rightarrow \dfrac{4\cos {{50}^{0}}}{\sin \left( {{90}^{0}}-{{50}^{0}} \right)} \\
\end{align}$

$\begin{align}
  & \Rightarrow \dfrac{4\cos {{50}^{0}}}{\cos {{50}^{0}}} \\
 & \Rightarrow 4 \\
\end{align}$

Thus, from the above calculation in which we have simplified the given term with the help of some trigonometric formula, we can say that the value of $\sqrt{3}\csc {{20}^{0}}-\sec {{20}^{0}}$ will be equal to 4.
Hence, (b) is the correct option.

Note:While using the formulae, one must make sure that the relations are properly used and there are no sign mistakes. Students get confused and write \[\cos \left( A+B \right)=\cos A\cdot \cos B-\sin A\cdot \sin B\] as \[\cos \left( A+B \right)=\cos A\cdot \cos B+\sin A\cdot \sin B\] , which is wrong. Such mistakes can lead to wrong answers and hence, should be avoided.