Prove the following trigonometric identities:

$s e c A (1 - s i n A) (s e c A + t a n A) = 1$

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Solution

Given,

L.H.S = secA(1 – sinA)(secA + tanA)

Here, secA = $\frac{1}{c o s A}$ and tanA = $\frac{s i n A}{c o s A}$

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

Take $\frac{1}{c o s A}$ outside from the brackets,

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

=> $\frac{1}{c o s^{2} A}$ (1 – sinA) (1+sinA)

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

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

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

=> $s e c^{2} A - t a n^{2} A = 1 = R H S$

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