Prove that $\sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A$.
Step-by-Step Solution
Key Concept: Multiply numerator and denominator inside the square root by $(1 + \sin A)$.
\text{LHS} = \sqrt{\dfrac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\dfrac{(1 + \sin A)^2}{1 - \sin^2 A}} = \sqrt{\dfrac{(1 + \sin A)^2}{\cos^2 A}}. [1.0 Mark]
$= \dfrac{1 + \sin A}{\cos A} = \dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS}$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Rationalising inside square root: 1.0 Mark
Simplifying to $\sec A + \tan A$: 1.0 Mark
Correct Answer: