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**(iv)** If a triangle ABC inscribed in a fixed circle be slightly varies in such a way as to have its varies aways on the circle, then show that $\frac{da}{\mathrm{cos}A}=\frac{db}{\mathrm{cos}B}=\frac{dc}{\mathrm{cos}C}=0$

There is a printing mistake in the question.

The correct question is If a triangle ABC inscribed in a fixed circle be slightly varies in such a way as to have its varies always on the circle, then show that $\frac{\mathrm{da}}{\mathrm{cosA}}+\frac{\mathrm{da}}{\mathrm{cos}{\displaystyle \mathrm{B}}}+\frac{\mathrm{da}}{\mathrm{cos}{\displaystyle \mathrm{C}}}=0$

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