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Question
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
Solution
L.H.S. = (1 + tan A . tan B)2 + (tan A – tan B)2
= 1 + tan2 A . tan2 B + 2 tan A . tan B + tan2 A + tan2 B – 2 tan A tan B
= 1 + tan2 A + tan2 B + tan2 A tan2 B
= sec2 A + tan2 B(1 + tan2 A)
= sec2 A + tan2 B sec2 A
= sec2 A(1 + tan2 B)
= sec2 A sec2 B = R.H.S.
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