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tan A=n tan B and sin A=m sin B.To prove  (cos A)^2=(m^2-1)/(n^2-1)

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You should remember that `tan A = sinA/cosA`  and `tan B = sinB/cos B` , hence, substituting `m sin B`  for `sin A ` in `tan A = n tan B`  yields:

`tan A = n tan B =>sinA/cosA = n(sinB/cos B) => (m sin B)/cosA = n (sinB/cos B)`

Reducing by duplicate factors yields:

`m/cos A = n/cos B => (m/cos A)^2 = (n/cos B)^2`

`(cos A)^2 = (m/n)^2*cos^2 B`

Substituting `1 - sin^2B ` for `cos^2 B`  yields:

`(cos A)^2 = (m/n)^2*(1 - sin^2B)`

Since `sin^2B = (sin^2A)/m^2`  yields:

`(cos A)^2 = (m/n)^2*(1 - (sin^2A)/m^2)`

Hence, evaluating `(cos A)^2`  under the given conditions yields `(cos A)^2 = (m/n)^2*(1 - (sin^2A)/m^2).`

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