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Question
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
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Solution
`cos A = (2sqrt(m))/(m + 1)` ...[Given]
We know that,
sin2A + cos2A = 1
∴ `sin^2A + ((2sqrt(m))/(m + 1))^2 = 1`
∴ `sin^2A + (4m)/(m + 1)^2 = 1`
∴ `sin^2A = 1 - (4m)/(m + 1)^2`
= `((m + 1)^2 - 4m)/(m + 1)^2`
= `(m^2 + 2m + 1 - 4m)/(m + 1)^2` ...[∵ (a + b)2 = a2 + 2ab + b2]
= `(m^2 - 2m + 1)/(m + 1)^2`
∴ `sin^2A = (m - 1)^2/(m + 1)^2` ...[∵ a2 – 2ab + b2 = (a – b)2]
∴ `sin A = (m - 1)/(m + 1)` ...[Taking square root of both sides]
Now, `"cosec" A = 1/(sin A)`
= `1/((m - 1)/(m + 1))`
∴ `"cosec" A = (m + 1)/(m - 1)`
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