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It is given that 3cot A = 4
Or,
Consider a right triangle ABC, right-angled at point B
If AB is 4k, then BC will be 3k, where k is a positive integer
In ΔABC,
(AC)2 = (AB)2 + (BC)2
= (4k)2 + (3k)2
= 16k2 + 9k2
= 25k2
AC = 5k
Therefore,
1 – tan2A/1 + tan2A = Cos2A – Sin2A
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