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If tanA=ntanB and sinA = msinB, then prove that cos^2A= m^-1 / n^+1
sinA = msinB....(1)tanA= ntanB(sinA/cosA) = n(sinB/cosB)....(2)substuting sinB value from equation (1)cosB = (n/m) cosA......(3)sin2A = m2sin2B1-cos2A = m2(1-cos2B)substituting equation (3)1-cos2A = m2[1 – ((n2/m2)cos2A)]1 – cos2A = m2 – n2cos2An2cos2A – cos2A = m2 – 1cos2A = (m2 – 1)/(n2 – 1)
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