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Sine Rule

Prove that ...

Question

Prove that $\frac{sin A + 1 - cos A}{sin A - 1 + cos A} = \frac{1 + sin A}{cos A}$

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Solution

$L H S = \frac{sin A + (1 - cos A)}{sin A - (1 - cos A)} = \frac{sin A + (1 - cos A)}{sin A - (1 - cos A)} \times \frac{sin A + (1 - cos A)}{sin A + (1 - cos A)} = \frac{{[sin A + (1 - cos A)]}^{2}}{sin^{2} A - {(1 - cos A)}^{2}} = \frac{sin^{2} A + 2 sin A (1 - cos A) + {(1 - cos A)}^{2}}{(1 - cos^{2} A) - {(1 - cos A)}^{2}} = \frac{(1 - cos^{2} A) + 2 sin A (1 - cos A) + {(1 - cos A)}^{2}}{(1 - cos^{2} A) - {(1 - cos A)}^{2}} = \frac{1 + cos A + 2 sin A + 1 - cos A}{1 + cos A - 1 + cos A} = \frac{2 (1 + sin A)}{2 cos A} = \frac{1 + sin A}{cos A} = R H S$

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

Q. Prove that :

\frac{1 + cos A + sin A}{1 + cos A - sin A} = \frac{1 + sinA}{cos A}

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

\frac{cos A}{1 + sin A} + \frac{1 + sin A}{cos A} = 2 sec A

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