Prove that: $\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A}$.
Step-by-Step Solution
Key Concept: LHS $= \dfrac{1 + 1/\cos A}{1/\cos A} = \cos A + 1 = 1 + \cos A$.<br>RHS $= \dfrac{1 - \cos^2 A}{1 - \cos A} = \dfrac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A$. LHS $=$ RHS.
LHS $= \dfrac{1 + 1/\cos A}{1/\cos A} = \cos A + 1$. [1.5 Marks]
RHS $= \dfrac{1 - \cos^2 A}{1 - \cos A} = \dfrac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A$. [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Simplifying LHS to $1 + \cos A$: 1.5 Marks
Simplifying RHS to $1 + \cos A$: 1.5 Marks
Correct Answer: