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Trigonometric Identities

Prove 1+cos...

Question

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

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Solution

Simplifying the LHS of $\frac{1 + cos A + sin A}{1 + cos A - sin A} = \frac{1 + sin A}{cos A}$ .
$\frac{1 + cos A + sin A}{1 + cos A - sin A} = \frac{1 + cos A + sin A}{1 + cos A - sin A} \times \frac{1 + cos A + sin A}{1 + cos A + sin A}$
$= \frac{{(1 + cos A + sin A)}^{2}}{{(1 + cos A)}^{2} - {sin}^{2} A}$
$= \frac{1 + {cos}^{2} A + {sin}^{2} A + 2 cos A + 2 sin A + 2 sin A cos A}{1 + {cos}^{2} A + 2 cos A - {sin}^{2} A}$
$= \frac{1 + 1 + 2 cos A + 2 sin A + 2 sin A cos A}{{cos}^{2} A + 2 cos A + 1 - {sin}^{2} A}$
$= \frac{2 + 2 cos A + 2 sin A + 2 sin A cos A}{{cos}^{2} A + 2 cos A + {cos}^{2} A}$
$= \frac{2 (1 + cos A) + 2 sin A (1 + cos A)}{2 cos A + 2 {cos}^{2} A}$
$= \frac{2 (1 + cos A) (1 + sin A)}{2 cos A (1 + cos A)}$
$= \frac{1 + sin A}{cos A}$
$= R H S$

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

Q. Prove the following identity.

\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}

Q. Prove the identity:

\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}{1 - 2 c o s^{2} A}

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}

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