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If tanA=ntanB and sinA=msinB, prove that...

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

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Given
`sinA=msinB`.....(1)
`tanA=ntanB`
`sinA/cosA​=nsinB/cosB`​....(2)
Substituting sinB from equation 1, we get
...
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