For a triangle ABC , the simplified from the expression ( cos 2A /a^2 - cos 2B /b^2 ) equals

Answers (1)

Solution:    

                 (fraccos 2Aa^2-fraccos 2Bb^2)=[frac1-2sin^2Aa^2-frac1-2sin^2Bb^2]

          Rightarrow        frac1a^2-frac1b^2-2(fracsin^2Aa^2-fracsin^2Bb^2)

         Rightarrow          frac1a^2-frac1b^2-2(frac14R^2-frac14R^2)

         Rightarrow          frac1a^2-frac1b^2.

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