Question

cos2A/a^2-cos2B/b^2=1/a^2-1/b^2

Answer 1

cos2Aa2−cos2Bb2=2cos2A−1a2−cos2B−1b2  [Using cos2A=2cos2A−1;cos2B=2cos2B−1 ]Now applying cos ruleCosA=b2+c2−a22bc;CosB=a2+c2−b22ac=2(b2+c2−a22bc)2−1a2−2(a2+c2−b22ac)2−1b2=(b2+c2−a2)2−2b2c22a2b2c2−(a2+c2−b2)2−2a2c22a2b2c2=b4+c4+2b2c2+a4−2a2(b2+c2)−2b2c2−[a4+c4+b4+2a2c2−2b2(a2+c2)−2a2c2]2a2b2c2=b4+c4+2b2c2+a4−2a2b2−2a2c2−2b2c2−[a4+c4+b4+2a2c2−2b2a2−2b2c2−2a2c2]2a2b2c2=2b2c2−2a2c22a2b2c2=1a2−1b2  Hence proved