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Question
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
Solution
Given,
m = a sec A + b tan A and n = a tan A + b sec A
m2 – n2 = (a sec A + b tan A)2 – ( a tan A + b sec A)2
= a2 sec2 A + b2 tan2 A + 2ab sec A tan A – (a2 tan2 A + b2 sec2 A + 2ab sec A tan A)
= sec2 A (a2 – b2) + tan2 A (b2 – a2)
= (a2 – b2) [sec2 A – tan2 A]
= (a2 – b2) [Since sec2 A – tan2 A = 1]
Hence, m2 – n2 = a2 – b2
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