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Trigonometric Identity- 1

1+cos A - sin...

Question

$\frac{1 + cos A - sin A}{1 + cos A + sin A} = sec A - tan A = \frac{1 - sin A}{cos A} = \sqrt{\frac{1 - sin A}{1 + sin A}}$

A

True

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B

False

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C

Ambiguous

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D

Data insufficient

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Solution

The correct option is B True
$\frac{1 + c o s A - s i n A}{1 + c o s A + s i n A}$
$= \frac{(1 + c o s A - s i n A)^{2}}{(1 + c o s A)^{2} - s i n^{2} A}$
$= \frac{1 + c o s^{2} A + s i n^{2} A - 2 c o s A s i n A - 2 s i n A + 2 c o s A}{(1 + c o s A)^{2} - (1 - c o s A) (1 + c o s A)}$
$= \frac{2 - 2 c o s A s i n A - 2 s i n A + 2 c o s A}{(1 + c o s A) (1 + c o s A - 1 + c o s A)}$
$= \frac{2 (1 + c o s A s i n A - s i n A + c o s A)}{2 c o s A (1 + c o s A)}$
$= \frac{1 + c o s A s i n A - s i n A + c o s A}{c o s A (1 + c o s A)}$
$= \frac{(1 + c o s A) - s i n A (1 + c o s A)}{c o s A (1 + c o s A)}$
$= s e c A - t a n A$
$= \frac{1 - s i n A}{c o s A}$
$= \sqrt{\frac{(1 - s i n A)^{2}}{c o s^{2} A}}$
$= \sqrt{\frac{(1 - s i n A)^{2}}{1 - s i n^{2} A}}$
$= \sqrt{\frac{(1 - s i n A)}{1 + s i n A}}$

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