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Question
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
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Solution
Given : sin θ + cos θ = m and secθ + cosecθ = n
Consider L.H.S. = n(m2 - 1) = (secθ + cosecθ)[(sinθ + cosθ)2 - 1]
= `(1/cosθ + 1/sinθ) [sin^2θ + cos^2θ + 2sinθcosθ - 1`]
= `((cosθ + sinθ)/(sinθcosθ)) (1 + 2sinθcosθ - 1)`
= `((cosθ + sinθ))/(sinθcosθ) (2 sinθ cosθ)`
= 2(sinθ + cosθ)
= 2m = R.H.S.
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