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