Question

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

Answer 1

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:

& \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{aligned}$$ Now, simplifying the given term. Then, $\sqrt{3}\csc {{20}^{0}}-\sec {{20}^{0}}$ Now, use equation (1) and equation (2) in the above term. Then, $\begin{aligned} & \sqrt{3}\csc {{20}^{0}}-\sec {{20}^{0}} \\\ & \Rightarrow \dfrac{\sqrt{3}}{\sin {{20}^{0}}}-\dfrac{1}{\cos {{20}^{0}}} \\\ & \Rightarrow \dfrac{\sqrt{3}\cos {{20}^{0}}-\sin {{20}^{0}}}{\sin {{20}^{0}}\cos {{20}^{0}}} \\\ & \Rightarrow \dfrac{2\left( \dfrac{\sqrt{3}}{2}\cos {{20}^{0}}-\dfrac{1}{2}\sin {{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\\ \end{aligned}$ Now, substituting $$\dfrac{\sqrt{3}}{2}=\cos {{30}^{0}}$$ from equation (3) and $$\dfrac{1}{2}=\sin {{30}^{0}}$$ from equation (4) in the above term. Then, $\begin{aligned} & \dfrac{2\left( \dfrac{\sqrt{3}}{2}\cos {{20}^{0}}-\dfrac{1}{2}\sin {{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\\ & \Rightarrow \dfrac{2\left( \cos {{30}^{0}}\cos {{20}^{0}}-\sin {{30}^{0}}\sin {{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\\ \end{aligned}$ Now, using the formula from the equation (5) with the value of $A={{30}^{0}}$ and $B={{20}^{0}}$ . Then, $\begin{aligned} & \dfrac{2\left( \cos {{30}^{0}}\cos {{20}^{0}}-\sin {{30}^{0}}\sin {{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\\ & \Rightarrow \dfrac{2\cos \left( {{30}^{0}}+{{20}^{0}} \right)}{\sin {{20}^{0}}\cos {{20}^{0}}} \\\ & \Rightarrow \dfrac{4\cos {{50}^{0}}}{2\sin {{20}^{0}}\cos {{20}^{0}}} \\\ \end{aligned}$ Now, using the formula from the equation (6) with the value of $\theta ={{20}^{0}}$ . Then, $\begin{aligned} & \dfrac{4\cos {{50}^{0}}}{2\sin {{20}^{0}}\cos {{20}^{0}}} \\\ & \Rightarrow \dfrac{4\cos {{50}^{0}}}{\sin {{40}^{0}}} \\\ \end{aligned}$ Now, substituting $\sin {{40}^{0}}=\cos {{50}^{0}}$ in the above term by using the formula from equation (7) with the value of $\theta ={{50}^{0}}$ . Then, $\begin{aligned} & \dfrac{4\cos {{50}^{0}}}{\sin {{40}^{0}}} \\\ & \Rightarrow \dfrac{4\cos {{50}^{0}}}{\sin \left( {{90}^{0}}-{{50}^{0}} \right)} \\\ \end{aligned}$ $\begin{aligned} & \Rightarrow \dfrac{4\cos {{50}^{0}}}{\cos {{50}^{0}}} \\\ & \Rightarrow 4 \\\ \end{aligned}$ Thus, from the above calculation in which we have simplified the given term with the help of some trigonometric formula, we can say that the value of $\sqrt{3}\csc {{20}^{0}}-\sec {{20}^{0}}$ will be equal to 4. Hence, (b) is the correct option. **Note** :While using the formulae, one must make sure that the relations are properly used and there are no sign mistakes. Students get confused and write $$\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.