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Question
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
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Solution
Given:
sin A + cos A = m and sec A + cosec A = n
Consider L.H.S = n (m2 – 1)
= `(secA + cosecA)[(sinA + cosA)^2 - 1]`
= `(1/cosA + 1/sinA)[sin^2A + cos^2A + 2sinAcosA - 1]`
= `((cosA + sinA)/(sinAcosA))(1 + 2sinAcosA - 1)`
= `((cosA + sinA))/(sinAcosA)(2sinAcosA)`
= 2(sin A + cos A)
= 2 m = R.H.S
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