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Range of Trigonometric Ratios from 0 to 90 Degrees

cos A 1+sin A...

Question

$\frac{cos A}{1 + sin A} + \frac{1 + sin A}{cos A}$ = ?

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Solution

$L H S = \frac{c o s A}{(1 + s i n A)} + \frac{(1 + s i n A)}{c o s A} = \frac{c o s ² A + (1 + s i n A) ²}{c o s A . (1 + s i n A)}$

$= c o s ² A + 1 + s i n ² A + 2 s i n A / c o s A . (1 + s i n A)$

use $s i n ² ϕ + c o s ² ϕ = 1$

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

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

$= \frac{2}{c o s A}$

$= 2 s e c$

$= R H S$

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

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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 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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