Question

Prove that $\frac{\sin 3A}{\cos A}+\frac{\cos 3A}{\sin A}=2\cot 2A$

Answer 1

$\frac{\sin 3A}{\cos A}+\frac{\cos 3A}{sinA}$

$\frac{3\sin A-4{\sin }^{3}A}{\cos A}+\frac{4{\cos }^{3}A-3\cos A}{sinA}$

As ${\sin }^{2}A=1-{\cos }^{2}A$ ${\cos }^{2}A=1-{\sin }^{2}A$

$\frac{3\sin A-4\sin A(1-{\cos }^{2}A)}{\cos A}+\frac{4\cos A(1-{\sin }^{2}A)-3\cos A}{sinA}$

$\frac{3\sin A-4\sin A+4\sin A{\cos }^{2}A}{\cos A}+\frac{4\cos A-4\cos A{\sin }^{2}A-3\cos A}{sinA}$

$\frac{-\sin A+4\sin A{\cos }^{2}A}{\cos A}+\frac{\cos A-4\cos A{\sin }^{2}A}{sinA}$

$-\tan A+4\sin A\cos A+\cot A-4\cos A\sin A$

$\cot A-\tan A$

$\cot A-\frac{1}{\cot A}$

$\frac{{\cot }^{2}A-1}{\cot A}$

$2\left(\frac{{\cot }^{2}A-1}{2\cot A}\right)$

$2\cot 2A$

(As $\cot 2A=\frac{{\cot }^{2}A-1}{2\cot A}$)

$LHS=2\cot 2A=RHS$

Hence proved.