Prove the following identities, where the angles involved are acute angles for which the expressions is defined.
sec A + tan A = $\sqrt{(\frac{1 + s i n A}{1 - s i n A})}$

[3 Marks]
[NCERT]
[Trigonometric Identities]

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Solution

Solution:

$\sqrt{(\frac{1 + s i n A}{1 - s i n A})} = R . H . S Dividing R.H.S by cos A : = \sqrt{(\frac{\frac{1}{c o s A} + \frac{s i n A}{c o s A}}{\frac{1}{c o s A} - \frac{s i n A}{c o s A}})}$
[1 Mark]
$= \sqrt{(\frac{s e c A + t a n A}{s e c A - t a n A})} = \sqrt{(\frac{s e c A + t a n A}{s e c A - t a n A}) \times (\frac{s e c A + t a n A}{s e c A + t a n A})}$
[1 Mark]
$= \frac{s e c A + t a n A}{\sqrt{s e c^{2} A - t a n^{2} A}} = \frac{s e c A + t a n A}{1} (∵ s e c^{2} A - t a n^{2} A = 1) = s e c A + t a n A = L . H . S \Rightarrow L . H . S = R . H . S P r o v e d .$
[1 Mark]

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