Question

prove that $\cot 3A=\frac{3\cot A-{\cot }^{3}A}{1-3{\cot }^{2}A}$

Answer 1

$\cot 3A=\frac{\cos 3A}{\sin 3A}⟶\left(1\right)$

We know that

$\cos 3A=4{\cos }^{3}A-3cosA⟶\left(2\right)$

$\sin 3A=3sinA-4{\sin }^{3}A⟶\left(3\right)$

Substituting $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$ we get

$\cot 3A=\frac{4{\cos }^{3}A-3cosA}{3sinA-4{\sin }^{3}A}$

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

$=cotA\left(\frac{{4\cos }^{2}A-3}{3-4{\sin }^{2}A}\right)$

$=cotA\left(\frac{4{\cos }^{2}A-3\times ({\sin }^{2}A+{\cos }^{2}A)}{3({\sin }^{2}A+{\cos }^{2}A)-4{\sin }^{2}A}\right)(\because {\sin }^{2}A+{\cos }^{2}A=1)$

$=cotA\left(\frac{4{\cos }^{2}A-3{\cos }^{2}A-3{\sin }^{2}A}{3{\cos }^{2}A+3{\sin }^{2}A-4{\sin }^{2}A}\right)$

$=cotA\left(\frac{{\cos }^{2}A-3{\sin }^{2}A}{3{\cos }^{2}A-{\sin }^{2}A}\right)$

$=cotA\times \frac{{\sin }^{2}A}{{\sin }^{2}A}\left(\frac{{\cos }^{2}A/{\sin }^{2}A-3}{3{\cos }^{2}A/{\sin }^{2}A-1}\right)$

$=cotA\left(\frac{{\cot }^{2}A-3}{3{\cot }^{2}A-1}\right)$

$=\frac{{\cot }^{3}A-3cotA}{3{\cot }^{2}A-1}$

$=\frac{3cotA-{\cot }^{3}A}{1-3{\cot }^{2}A}$

Hence proved.