Question

# Prove that $\dfrac{1}{\sin {{10}^{o}}}-\dfrac{\sqrt{3}}{\cos {{10}^{o}}}=4$.

Hint: Use formula $\sin 2\theta =2\sin \theta \cos \theta$and express equation in form of
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$.

We have to prove that $\dfrac{1}{\sin {{10}^{o}}}-\dfrac{\sqrt{3}}{\cos {{10}^{o}}}=4....\left( i \right)$
Taking left hand side of equation $\left( i \right)$
$\Rightarrow \dfrac{1}{\sin {{10}^{o}}}-\dfrac{\sqrt{3}}{\cos {{10}^{o}}}$
$=\dfrac{\cos {{10}^{o}}-\left( \sin {{10}^{o}} \right).\sqrt{3}}{\sin {{10}^{o}}.\cos {{10}^{o}}}$
Now, we will multiply numerator and denominator by $2$.
We get, $\dfrac{2\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{2\sin {{10}^{o}}.\cos {{10}^{o}}}....\left( ii \right)$
Now, we know that $2\sin \theta \cos \theta =\sin 2\theta$
Therefore, $2\sin {{10}^{o}}.\cos {{10}^{o}}=\sin 2\left( {{10}^{o}} \right)$$=\sin {{20}^{o}}$
We will put the value of $2\sin {{10}^{o}}\cos {{10}^{o}}$in equation $\left( ii \right)$.
We get $\dfrac{2\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}$
We will multiply and divide numerator by $2$.
We get, $\dfrac{2.2.\dfrac{1}{2}\left( \cos {{10}^{o}}-\sqrt{3}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}$
By rearranging the equation and taking $\dfrac{1}{2}$inside the bracket.
We get, $\dfrac{4\left( \dfrac{1}{2}\cos {{10}^{o}}-\dfrac{\sqrt{3}}{2}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}...\left( iii \right)$
Now, we know that $\sin {{30}^{o}}=\dfrac{1}{2}$and $\cos {{30}^{o}}=\dfrac{\sqrt{3}}{2}$.
We put these values in equation $\left( iii \right)$.
We get $\dfrac{4\left( \sin {{30}^{o}}\cos {{10}^{o}}-\cos {{30}^{o}}\sin {{10}^{o}} \right)}{\sin {{20}^{o}}}....\left( iv \right)$
We know that $\sin A\cos B-\cos A\sin B=\sin \left( A-B \right)$
Therefore, $\sin {{30}^{o}}\cos {{10}^{o}}-\cos {{30}^{o}}\sin {{10}^{o}}=\sin \left( {{30}^{o}}-{{10}^{o}} \right)=\sin {{20}^{o}}$
Putting these values in equation $\left( iv \right)$
We get, $\dfrac{4\sin {{20}^{o}}}{\sin {{20}^{o}}}$
$=4$$=$Right hand side or RHS
Therefore, $LHS=RHS$[Hence Proved]

Note: Some students try to use $\dfrac{1}{\sin {{10}^{o}}}=\operatorname{cosec}{{10}^{o}}$ and
$\dfrac{1}{\cos {{10}^{o}}}=\sec {{10}^{o}}$in question to get away with fractions, but this actually
creates confusion and makes solution lengthy.