Question

What is the value of $\dfrac{1}{{\sin {{10}^ \circ }}} - \dfrac{{\sqrt 3 }}{{\cos {{10}^ \circ }}} = $
A. $0$
B. $1$
C. $2$
D. $4$

Answer 1

Hint: In order to solve this type of question, first we will consider the given equation. Then, we will simplify it by using multiplication and division. Next, we will apply suitable trigonometric identities ($\left[ {\because 2\sin \theta \cos \theta = \sin 2\theta } \right]$ and $\left[ {\because \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B} \right]$) in the equation formed and simplify it further in order to get the desired answer.

Formula used: $\left[ {\because 2\sin \theta \cos \theta = \sin 2\theta } \right]$
$\left[ {\because \sin {{30}^ \circ } = \dfrac{1}{2}} \right]$
$\left[ {\because \cos {{30}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right]$
$\left[ {\because \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B} \right]$

Complete step by step solution:
Consider,
$\dfrac{1}{{\sin {{10}^ \circ }}} - \dfrac{{\sqrt 3 }}{{\cos {{10}^ \circ }}}$
\[ = \dfrac{{\cos {{10}^ \circ } - \sqrt 3 \sin {{10}^ \circ }}}{{\sin {{10}^ \circ }\cos {{10}^ \circ }}}\]
Multiply the numerator and denominator by $2,$
\[ = \dfrac{{2\left( {\cos {{10}^ \circ } - \sqrt 3 \sin {{10}^ \circ }} \right)}}{{2\left( {\sin {{10}^ \circ }\cos {{10}^ \circ }} \right)}}\]
Multiply and divide the numerator by $2,$
\[ = \dfrac{{2 \times 2 \times \dfrac{1}{2}\left( {\cos {{10}^ \circ } - \sqrt 3 \sin {{10}^ \circ }} \right)}}{{\sin {{20}^ \circ }}}\] $\left[ {\because 2\sin \theta \cos \theta = \sin 2\theta } \right]$
\[ = \dfrac{{4\left( {\dfrac{1}{2}\cos {{10}^ \circ } - \dfrac{{\sqrt 3 }}{2}\sin {{10}^ \circ }} \right)}}{{\sin {{20}^ \circ }}}\]
Simplifying it,
\[ = \dfrac{{4\left( {\sin {{30}^ \circ }\cos {{10}^ \circ } - \cos {{30}^ \circ }\sin {{10}^ \circ }} \right)}}{{\sin {{20}^ \circ }}}\] $\left[ {\because \sin {{30}^ \circ } = \dfrac{1}{2}} \right],\left[ {\because \cos {{30}^ \circ } = \dfrac{{\sqrt 3 }}{2}} \right]$
\[ = \dfrac{{4\sin \left( {{{30}^ \circ } - {{10}^ \circ }} \right)}}{{\sin {{20}^ \circ }}}\] $\left[ {\because \sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B} \right]$
Solving it,
\[ = \dfrac{{4\sin {{20}^ \circ }}}{{\sin {{20}^ \circ }}}\]
$ = 4$
Therefore the correct option is D.

Note: To solve this question, trigonometric equations as well as the value of sine and cosine functions at particular angles should be remembered. Trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles.