(1+tanA×tanB)+(tanA-tanB)²=sec²A×sec²B prove that
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Answer:
LHS = (1 + tan a 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
= 1. (1 + tan²b) + tan²a (tan²b + 1)
= (1 + tan²b) + (1 + tan²a)
= sec²b× sec²a
= sec2a sec2b
= RHS
Step-by-step explanation:
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