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