Question

If tan A = n tan B and sin A = m sin B , prove that  `cos^2 A = ((m^2-1))/((n^2 - 1))`

Answer 1

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))`