Prove that cot (π/4 – 2 cot-1 3) = 7

Re-writing the given,

$\frac{\pi}{4}-2 \cot ^{-1} 3=\cot ^{-1} 7$

$2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}-\tan ^{-1} \frac{1}{7}$

$2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}$

Now, $2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1} \frac{2 / 3}{1-(1 / 3)^{2}}+\tan ^{-1} \frac{1}{7}\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$

$=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \cdot \frac{1}{7}}$

$=\tan ^{-1} \frac{(21+4) / 28}{\left(28^{4}-3\right) / 28}=\tan ^{-1} \frac{25}{25}=\tan ^{-1} 1=\frac{\pi}{4}$

$\frac{\pi}{4}-2 \cot ^{-1} 3=\cot ^{-1} 7$

$2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}-\tan ^{-1} \frac{1}{7}$

$2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}$

Now, $2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1} \frac{2 / 3}{1-(1 / 3)^{2}}+\tan ^{-1} \frac{1}{7}\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$

$=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \cdot \frac{1}{7}}$

$=\tan ^{-1} \frac{(21+4) / 28}{\left(28^{4}-3\right) / 28}=\tan ^{-1} \frac{25}{25}=\tan ^{-1} 1=\frac{\pi}{4}$

L.H.S = R.H.S

– Hence Proved