Question

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

Answer 1

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

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

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