1.

If tanA=ntanB and sinA=msinB, prove that `cos^2A=[m^2-1]/[n^2-1]`

Answer» `tanA = ntanB=> n = tanA/tanB`
`sinA = msinB => m = sinA/sinB`
`R.H.S. = (m^2-1)/(n^2-1) = (sin^2A/sin^2B-1)/(tan^2A/tan^2B-1)`
`=((sin^2A-sin^2B)/sin^2B)/((tan^2A-tan^2B)/tan^2B)`
`=((sin^2A-sin^2B)/sin^2B)/((tan^2A-tan^2B)/(sin^2B/cos^2B))`
`=(sin^2A-sin^2B)/(cos^2B(tan^2A-tan^2B))`
`=(1-cos^2A-(1-cos^2B))/(cos^2B(sec^2A-1-(sec^2B-1)))`
`=(cos^2B-cos^2A)/(cos^2B(sec^2A-sec^2B))`
`=(cos^2B-cos^2A)/(cos^2B(1/cos^2A-1/cos^2B)`
`=(cos^2Acos^2B(cos^2B-cos^2A))/(cos^2B(cos^2B-cos^2A)`
`=cos^2A = L.H.S.`


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