Question

Prove that $\cos \left[2{\tan }^{-1}\frac{1}{7}\right]=\sin \left[4{\tan }^{-1}\frac{1}{3}\right]$.

Answer 1

L.H.S $=cos\left(2ta{n}^{-1}1/7\right)$

$=cos\left(2a\right)[a=ta{n}^{-1}1/7.]$

$=\frac{1-ta{n}^{2}a}{1+ta{n}^{2}a}|\therefore a=ta{n}^{}-11/7\Rightarrow tana=1/7|$

$\frac{1-(1/7{)}^2}{1+(1/7{)}^{}2}$

$=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}$

$=\frac{\frac{49-1}{49}}{\frac{49+1}{49}}$

$=\frac{48}{50}$

$=\frac{24}{25}$

R.H.S $=sin\left[4ta{n}^{-1}\frac{1}{3}\right]=sin\left[4b\right][letb=ta{n}^{-1}\frac{1}{3}\Rightarrow tanb=\frac{1}{3}]$

= 2sin 2b cos 2b.

$=2\left(\frac{2tanb}{1+ta{n}^{2}b}\right)\left(\frac{1-ta{n}^{2}b}{1+ta{n}^{2}b}\right)$

$=2\left(\frac{2.\frac{1}{3}}{1+(\frac{1}{3}{)}^2}\right)\left(\frac{1-(\frac{1}{3}{)}^2}{1+(\frac{1}{3}{)}^2}\right)$

$=2.\left(\frac{\frac{2}{3}}{\frac{9+1}{3^2}}\right)\left(\frac{\frac{9-1}{9}}{\frac{9+1}{9}}\right)$

$=2.\frac{6}{10}.\frac{8}{10}$

$=\frac{24}{25}$

$ \therefore L.H.S =R.H.S.