If 3 cot A = 4, check whether \(1 – tan^2A\over 1 + tan^2A\) = \(cos^2A – sin^2A\) or not.

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 :

\(1 – tan^2A\over 1 + tan^2A\) = \(cos^2A – sin^2A\)

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