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Question
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
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Solution
L.H.S. = (1 + tanA . tan B)2 + (tan A – tan B)2
= 1 + tan2A . tan2B + 2 tan A . tan B + tan2A + tan2B – 2 tan A tan B
= 1 + tan2A + tan2B + tan2A + tan2B
= sec2A + tan2 B(1 + tan2A)
= sec2A + tan2 B sec2 A
= sec2A(1 + tan2B)
= sec2A sec2B
= R.H.S.
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