Prove that: $\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$.
Step-by-Step Solution
Key Concept: Divide numerator and denominator by $\sin A$: $\dfrac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A} = \dfrac{(\cot A + \csc A) - (\csc^2 A - \cot^2 A)}{\cot A - \csc A + 1} = \dfrac{(\csc A + \cot A)(1 - \csc A + \cot A)}{\cot A - \csc A + 1} = \csc A + \cot A$.
Divide by $\sin A$: LHS $= \dfrac{\cot A + \csc A - 1}{\cot A - \csc A + 1}$. [1.0 Mark]
Substitute $1 = \csc^2 A - \cot^2 A$: $= \dfrac{(\cot A + \csc A) - (\csc A - \cot A)(\csc A + \cot A)}{\cot A - \csc A + 1}$. [1.0 Mark]
$= \dfrac{(\csc A + \cot A)(1 - \csc A + \cot A)}{\cot A - \csc A + 1} = \csc A + \cot A = $ RHS. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Dividing by $\sin A$: 1.0 Mark
Substituting $1 = \csc^2 A - \cot^2 A$: 1.0 Mark
Factoring to get $\csc A + \cot A$: 1.0 Mark
Correct Answer: