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Chain Rule of Differentiation

Prove that ...

Question

Prove that $\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{s i n^{2} A - c o s^{2} A}$

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Solution

To prove:- $\frac{sin A + cos A}{sin A - cos A} + \frac{sin A - cos A}{sin A + cos A} = \frac{2}{{sin}^{2} A - {cos}^{2} A}$
Proof: $\frac{(sin A + cos A)^{2} + (sin A - cos A)^{2}}{(sin A - cos A) (sin A + cos A)}$
$= \frac{{sin}^{2} A + {cos}^{2} A + 2 sin A \cdot cos A + {sin}^{2} A + {cos}^{2} A - 2 sin A \cdot cos A}{{sin}^{2} A - {cos}^{2} A}$
$= \frac{1 + 1}{{sin}^{2} A - {cos}^{2} A} = \frac{2}{{sin}^{2} A - {cos}^{2} A} =$ R.H.S.
$[∵ {sin}^{2} A + {cos}^{2} A = 1]$ .

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\frac{s i n A + c o s A}{s i n A - c o s A} + \frac{s i n A - c o s A}{s i n A + c o s A} = \frac{2}{2 s i n^{2} A - 1}

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