If tanA=n tanB and sin A =m sin B, prove that cos^2A=m^2-n^2 - 1
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tanA=n tanB
sinA/cosA=n sinB/cosB
m sinB/cosA=n sinB/cosB
m cosB =n cosA
squaring both sides
m^2 cos^2B=n^2 cos^2A
cos^2A=m^2 cos^2B/n^2
=(m/n)^2 ×(1- sin^2B)
=(m/n)^2×(1-sin^2A/m^2)
= (m/n)^2×(m^2-sin^2A)/m^2
=n^2×(m^2-sin^2A)
m^2-n^2-1
=n^2×(cosA/cosB)^2-n^2-1
=(n^2 cos^2A-n^2cos^2B-1)/cos^B
=n^2(cos^2A-cos^2B
sinA/cosA=n sinB/cosB
m sinB/cosA=n sinB/cosB
m cosB =n cosA
squaring both sides
m^2 cos^2B=n^2 cos^2A
cos^2A=m^2 cos^2B/n^2
=(m/n)^2 ×(1- sin^2B)
=(m/n)^2×(1-sin^2A/m^2)
= (m/n)^2×(m^2-sin^2A)/m^2
=n^2×(m^2-sin^2A)
m^2-n^2-1
=n^2×(cosA/cosB)^2-n^2-1
=(n^2 cos^2A-n^2cos^2B-1)/cos^B
=n^2(cos^2A-cos^2B
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