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Basic Trigonometric Identities

Prove that : ...

Question

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

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Solution

We have, $\frac{1 + c o s A + s i n A}{1 + c s o s A - s i n A}$

$= \frac{1 + c o s A + s i n A}{1 + c o s A - s i n A} \times \frac{(1 + c o s A) + s i n A}{(1 + c o s A) + s i n A}$

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

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

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

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

$\frac{1 + 1 + 2 c o s A + 2 s i n A + 2 s i n A c o s A}{2 c o s A (1 + c o s A)}$

$\frac{2 + 2 c o s A + 2 s i n A + 2 s i n A c o s A}{2 c o s A (1 + c o s A)}$

$\frac{1 + c o s A + s i n A + s i n A c o s A}{c o s A (1 + c o s A)}$

$= \frac{1 + s i n A + c o s A (1 + s i n A)}{c o s A (1 + c o s A)}$

$= \frac{(1 + s i n A) (1 + c o s A)}{c o s A (1 + c o s A)}$

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

Hence proved.

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Similar questions

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}

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

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

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