Question

Find the value of \[\sqrt{3}\operatorname{cosec}{{20}^{o}}-\sec {{20}^{o}}\].

Answer 1

Hint: First of all convert the whole equation in terms of \[\sin \theta \] and \[\cos \theta \] by using \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] and \[sec\theta =\dfrac{1}{\cos \theta }\]. Then multiply the numerator and denominator by \[2.\dfrac{1}{2}\] and use \[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\] and \[\sin 2A=2\sin A\cos A\] to further solve the question.

Complete step-by-step answer:
Here, we have to find the value of the expression \[\sqrt{3}\operatorname{cosec}{{20}^{o}}-\sec {{20}^{o}}\]. Let us consider the expression given in the question
\[E=\sqrt{3}\operatorname{cosec}{{20}^{o}}-\sec {{20}^{o}}\]

Let us first convert the given expression in terms of \[\sin \theta \] and \[\cos \theta \]. We know that \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] and \[sec\theta =\dfrac{1}{\cos \theta }\]. By using this in the above expression, we get,
\[\begin{align}
  & E=\dfrac{\sqrt{3}}{\sin {{20}^{o}}}-\dfrac{1}{\cos {{20}^{o}}} \\
 & E=\dfrac{\left( \sqrt{3}\cos {{20}^{o}}-\sin {{20}^{o}} \right)}{\left( \sin {{20}^{o}} \right).\left( \cos {{20}^{o}} \right)} \\
\end{align}\]
By multiplying \[2\times \dfrac{1}{2}\] on both numerator and denominator of the above expression, we get,
\[E=\dfrac{2\times \dfrac{1}{2}\left( \sqrt{3}\cos {{20}^{o}}-\sin {{20}^{o}} \right)}{2\times \dfrac{1}{2}\left( \sin {{20}^{o}} \right).\left( \cos {{20}^{o}} \right)}\]
We can also write the above expression as,
\[E=\dfrac{2\left( \dfrac{\sqrt{3}}{2}\cos {{20}^{o}}-\dfrac{1}{2}\sin {{20}^{o}} \right)}{\dfrac{1}{2}\left( 2\sin {{20}^{o}}.\cos {{20}^{o}} \right)}\]
\[\Rightarrow E=\dfrac{4\left( \dfrac{\sqrt{3}}{2}\cos {{20}^{o}}-\dfrac{1}{2}\sin {{20}^{o}} \right)}{\left( 2\sin {{20}^{o}}.\cos {{20}^{o}} \right)}\]
We know that \[\sin {{60}^{o}}=\dfrac{\sqrt{3}}{2}\] and \[\cos {{60}^{o}}=\dfrac{1}{2}\]. By using this in the above expression, we get,
\[E=\dfrac{4\left( \sin {{60}^{o}}\cos {{20}^{o}}-\cos {{60}^{o}}\sin {{20}^{o}} \right)}{\left( 2\sin {{20}^{o}}.\cos {{20}^{o}} \right)}\]
We know that \[\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)\]. By using it in the numerator of the above expression, we get,
\[E=\dfrac{4\left( \sin \left( {{60}^{o}}-{{20}^{o}} \right) \right)}{2\sin {{20}^{o}}\cos {{20}^{o}}}\]
So, we get, \[E=\dfrac{4\left( \sin {{40}^{o}} \right)}{2\sin {{20}^{o}}\cos {{20}^{o}}}\]

Now, we know that \[2\sin A\cos A=\sin 2A\]. By using this in the denominator of the above expression, we get,
\[E=\dfrac{4\left( \sin {{40}^{o}} \right)}{\sin \left( 2\left( {{20}^{o}} \right) \right)}\]
\[\Rightarrow E=\dfrac{4\left( \sin {{40}^{o}} \right)}{\left( \sin {{40}^{o}} \right)}\]
By canceling the like terms from the above expression, we get E = 4.
So, we get the value of \[\sqrt{3}\operatorname{cosec}{{20}^{o}}-\sec {{20}^{o}}=4\].

Note: Students must remember the general trigonometric formulas like \[\sin \left( A\pm B \right)=\sin A\cos B\pm \cos A\sin B\] and \[\sin 2A=2\sin A\cos A\]. Also, students should keep in mind the values of sine and cosine at general angles like \[{{0}^{o}},{{30}^{o}},{{60}^{o}},{{45}^{o}}\text{ and }{{90}^{o}}\]. Always try to convert the given expression into some trigonometric formulas in these types of questions.