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Question

If tanA=ntanB and sinA=msinB ,prove that cos2A =m2 -1/n2 - 1

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Solution

Tan A = n Tan B ---(1)
Sin A = m Sin B --- (2)
(2) ÷ (1) gives: CosA = (m/n) CosB --(3)

From (2)
Cos² A = 1 - m² Sin²B . [Cos² A+Sin² A=1]

= 1 - m² [1 - cos²B]

= 1 - m² + m² Cos²B

= 1 - m² + n² Cos²A using (3)

=> (n²-1) Cos²A = m² -1
=> Cos²A = (m² - 1) / (n² - 1)


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