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Trigonometric Ratios of Common Angles

1-sin A cos A...

Question

$\frac{1 - s i n A c o s A}{c o s A (s e c A - c o s e c A)} . \frac{s i n^{2} A - c o s^{2} A}{s i n^{3} A + c o s^{3} A} = s i n A$

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Solution

$L H S = \frac{1 - s i n A c o s A}{c o s A (s e c A - c o s e c A)} . \frac{s i n^{2} A - c o s^{2} A}{s i n^{3} A + c o s^{3} A}$

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

$[\begin{matrix} U s i n g a^{2} - b^{2} = (a - b) (a + b) a n d a^{3} + b^{3} - (a + b) (a^{2} + b^{2} - a b) \end{matrix}]$

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

$= \frac{c o s A s i n A}{c o s A} = s i n A = R H S$

Hence proved.

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