Question

Prove that: $\cot (\frac{\pi }{2}-2{\cot }^{-1}3)=\frac{3}{4}$

Answer 1

L.H.S

$\cot (\frac{\pi }{2}-2{\cot }^{-1}3)$

We know that

$\cot (\frac{\pi }{2}-\theta )=\tan \theta$

Therefore,

$\Rightarrow \tan \left(2{\cot }^{-1}3\right)$

$\Rightarrow \tan \left(2{\tan }^{-1}\left(\frac{1}{3}\right)\right)[\because {\cot }^{-1}x={\tan }^{-1}\left(\frac{1}{x}\right)]$

$\Rightarrow \tan \left({\tan }^{-1}\left(\frac{2\times \frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}\right)\right)[2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)]$

$\Rightarrow \frac{\frac{2}{3}}{1-\frac{1}{9}}$

$\Rightarrow \frac{\frac{2}{3}}{\frac{8}{9}}$

$\Rightarrow \frac{2\times 9}{3\times 8}$

$\Rightarrow \frac{3}{4}$

Hence, this is the answer.