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Question
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Solution
We have tanA = n tan B
⇒ `cot B = n/tanA ........(i)`
Again , sin A = m sin B
` ⇒ cosec B = m/ sin A ........(ii) `
Squaring (i) and ( ii) and subtracting (ii) from (i) , We get
`⇒ (m^2)/(sin^2 A) - (n^2 )/(tan^2 A) = cosec ^2 B - cot^2 B`
`⇒ (m^2 )/(sin^2 A )-(n^2 cos )/(sin^2 A)m=1`
`⇒m^2 - n^2 cos^2 A = sin^2 A`
`⇒ m^2 - n^2 cos^2 A =1- cos^2 A`
`⇒ n^2 cos^2 A- cos^2 A = m^2 -1`
`⇒cos^2 A (n^2 -1) = (m^2 -1)`
`⇒ cos^2 A = ((m^2 -1))/((n^2 -1))`
∴` cos^2 A = ((m^2 -1))/((n^2 -1))`
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