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| 1. |
If tanA=ntanB, sinA=msinB then prove that cos2A=m2 - 1/n2-1 |
| Answer» SinA = m Sin B , Sin A/ Sin B = m and\xa0tan A= n tan B, tan A/tan B= n or Sin A/ cos A/Sin B/cos B= n , SinA Cos B / Cos A. Sin B= nm2\xa0-1/ n2\xa0-1 = (Sin A / Sin B )2\xa0-1/(Sin ACos B /Cos A Sin B)2\xa0-1 = Sin2A/ Sin2B -1/ Sin2ACos2B/Cos2ASin2B -1 = Sin2A-Sin2B/ Sin2B/Sin2ACos2B - Cos2A Sin2B/ Cos2A Sin2B = Sin2A - Sin2B x Cos2A Sin2B/Sin2B (Sin2A Cos2B - Cos2A Sin2B) = Sin2A - Sin2B x Cos2A/ Sin2A(1- sin2B) - (1-Sin2A) Sin2B = (Sin2A - Sin\xa02B ) x Cos2A /Sin2A - Sin2A Sin2B - Sin2B +Sin2A Sin2B = (Sin2A - Sin2B) x Cos2A/ ( Sin2A - Sin2B) = Cos\xa02A | |