Solution

We have to prove:

\[ \cot\left(\frac{\pi}{4} – 2\cot^{-1}3\right) = 7 \]

Taking cot inverse on both sides:

\[ \frac{\pi}{4} – 2\cot^{-1}3 = \cot^{-1}7 \]

\[ 2\cot^{-1}3 = \frac{\pi}{4} – \cot^{-1}7 \]

Since \(\cot^{-1}x = \tan^{-1}\left(\frac{1}{x}\right)\),

\[ 2\tan^{-1}\frac{1}{3} = \frac{\pi}{4} – \tan^{-1}\frac{1}{7} \]

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

Using identity: \[ 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1 – x^2}\right) \]

\[ \tan^{-1}\frac{2/3}{1 – (1/3)^2} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4} \]

\[ \tan^{-1}\frac{2/3}{8/9} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4} \]

\[ \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4} \]

\[ \tan^{-1}\left(\frac{\frac{3}{4} + \frac{1}{7}}{1 – \frac{3}{4}\cdot\frac{1}{7}}\right) = \frac{\pi}{4} \]

\[ \tan^{-1}\frac{25}{25} = \frac{\pi}{4} \]

\[ \tan^{-1}(1) = \frac{\pi}{4} \]

\[ \boxed{\text{LHS} = \text{RHS}} \]