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Question
Prove the following:
(1 + tanA · tanB)2 + (tanA − tanB)2 = sec2A · sec2B
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Solution
L.H.S. = (1 + tan A · tan B)2 + (tan A − tan B)2
= 1 + 2 tan A · tan B + tan2A tan2B + tan2A – 2 tan A · tan B + tan2B
= 1 + tan2A + tan2B + tan2A tan2B
= 1(1 + tan2A) + tan2B(1 + tan2A)
= (1 + tan2A) (1 + tan2B)
= sec2A sec2B
= R.H.S.
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