Show that cos (2 tan-1 1/7) = sin (4 tan-1 1/3).

Solution:

Taking L.H.S, we have

L.H.S. $=\cos \left(2 \tan ^{-1} \frac{1}{7}\right)$

$=\cos \left(\cos ^{-1} \frac{1-\left(\frac{1}{7}\right)^{2}}{1+\left(\frac{1}{7}\right)^{2}}\right) \quad\left(\because 2 \tan ^{-1} x=\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right)$

$=\cos \left(\cos ^{-1} \frac{48 / 49}{50 / 49}\right)$

$=\cos \left(\cos ^{-1} \frac{24}{25}\right)=\frac{24}{25} \quad\left(\because \cos \left(\cos ^{-1} x\right)=x, x \in[-1,1]\right)$

R.H.S. $=\sin \left(4 \tan ^{-1} \frac{1}{3}\right)$

$=\sin \left(2\left(2 \tan ^{-1} \frac{1}{3}\right)\right)$

$=\sin \left(2\left(\tan ^{-1} \frac{2 \cdot \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}}\right)\right) \quad\left(\because 2 \tan ^{-1} x \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right)$

$=\sin \left(2 \tan ^{-1} \frac{2 / 3}{8 / 9}\right)$

$=\sin \left(2 \tan ^{-1} \frac{3}{4}\right)$

$=\sin \left(\sin ^{-1} \frac{2 \times \frac{3}{4}}{1+\left(\frac{3}{4}\right)^{2}}\right) \quad\left(\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}\right)$

$=\sin \left(\sin ^{-1} \frac{3 / 2}{25 / 16}\right)$

$=\sin \left(\sin ^{-1} \frac{24}{25}\right)=\frac{24}{25} \quad\left(\because \sin \left(\sin ^{-1} x\right)=x, x \in[-1,1]\right)$

L.H.S. $=$ R.H.S.

Thus, L.H.S = R.H.S

– Hence proved