Question

if 3tanA=4 cheak weather 1-tan^2A/1+tan^2A=cos^2asin^2A

Answer 1

Answer:

Let ΔABC in which ∠B = 90º, A/q, cot A = AB/BC = 4/3                                                Let AB = 4k and BC = 3k, where k is a positive real number.                                           By Pythagoras theorem in ΔABC we get. AC2 = AB2 + BC2 AC2 = (4k)2 + (3k)2 AC2 = 16k2 + 9k2 AC2 = 25k2 AC = 5k tan A = BC/AB = 3/4                                                                                                                                                                                       sin A = BC/AC = 3/5 cos A = AB/AC = 4/5                                                                             L.H.S. = (1-tan2A)/(1+tan2A) = 1- (3/4)2/1+ (3/4)2 = (19/16)/(1+ 9/16) = (16-9)/(16+9) = 7/25                                                                                                                    R.H.S. = cos2A – sin2A = (4/5)2 - (3/4)2 = (16/25) - (9/25) = 7/25 R.H.S. = L.H.S. Hence,  (1-tan2A)/(1+tan2A) = cos2A – sin2A

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