if cosA / cosB = M and cosA / sinB = N, prove that ( m² + n² ) cos² B = n²
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L.H.S= (m2+n2)cos2b
= {(cosa/cosb)2 +(cosa/sinb)2}cos2b
= {(cos2a/cos2b) +(cos2a/sin2b)}cos2b
= (cos2a.sin2b +cos2a.cos2b / cos2b.sin2b)cos2b
= cos2a(sin2b+cos2b) / sin2b
= cos2a / sin2b
R.H.S= n2
=(cosa/sinb)2
= cos2a / sin2b
Therefore, L.H.S=R.H.S
= {(cosa/cosb)2 +(cosa/sinb)2}cos2b
= {(cos2a/cos2b) +(cos2a/sin2b)}cos2b
= (cos2a.sin2b +cos2a.cos2b / cos2b.sin2b)cos2b
= cos2a(sin2b+cos2b) / sin2b
= cos2a / sin2b
R.H.S= n2
=(cosa/sinb)2
= cos2a / sin2b
Therefore, L.H.S=R.H.S
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