Solution :
We have, 3 cot A = 4 \(\implies\) cot A = \(4\over 3\) = \(AB\over BC\) 
Let AB = 4k, then BC = 3k
By Pythagoras Theorem,
\({AC}^2\) = \({AB}^2\) + \({BC}^2\)
\(\implies\) \({AC}^2\) = \(25k^2\)
\(\implies\) AC = 5k
Thus, tan A = \(BC\over AB\) = \(3k\over 4k\) = \(3\over 4\)
sin A = \(BC\over AC\) = \(3k\over 5k\) = \(3\over 5\)
cos A = \(AB\over AC\) = \(4k\over 5k\) = \(4\over 5\)
Now, \(1 – tan^2A\over 1 + tan^2A\) = \(16 – 9\over 16 + 9\) = \(7\over 25\) …………(1)
and \(cos^2A – sin^2A\) = \(7\over 25\) ………….(2)
From (1) and (2), we have :
\(1 – tan^2A\over 1 + tan^2A\) = \(cos^2A – sin^2A\)
