Prove that $\dfrac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \dfrac{1}{\sec A - \tan A}$ using the identity $\sec^2 A = 1 + \tan^2 A$.
Step-by-Step Solution
Key Concept: Divide numerator and denominator by $\cos A$ to convert to $\tan A$ and $\sec A$, then replace $1 = \sec^2 A - \tan^2 A$.
Stepwise Solution:
Divide numerator and denominator by $\cos A$:
\text{LHS} = \dfrac{\tan A - 1 + \sec A}{\tan A + 1 - \sec A} = \dfrac{(\tan A + \sec A) - 1}{(\tan A - \sec A) + 1}. [1.5 Marks]
Substitute $1 = \sec^2 A - \tan^2 A = (\sec A - \tan A)(\sec A + \tan A)$ in numerator:
$= \dfrac{(\tan A + \sec A) - (\sec A - \tan A)(\sec A + \tan A)}{(\tan A - \sec A) + 1}. [1.5 Marks]
Factor out $(\sec A + \tan A)$:
$= \dfrac{(\sec A + \tan A)[1 - (\sec A - \tan A)]}{\tan A - \sec A + 1} = \dfrac{(\sec A + \tan A)(1 - \sec A + \tan A)}{1 - \sec A + \tan A} = \sec A + \tan A$. [1.0 Mark]
Multiply and divide by $(\sec A - \tan A)$:
$= \dfrac{(\sec A + \tan A)(\sec A - \tan A)}{\sec A - \tan A} = \dfrac{\sec^2 A - \tan^2 A}{\sec A - \tan A} = \dfrac{1}{\sec A - \tan A} = \text{RHS}$. Proved! [1.0 Mark]
Marking Scheme:
• Dividing by $\cos A$ to get terms in $\sec A$ and $\tan A$: 1.5 Marks
• Substituting $1 = \sec^2 A - \tan^2 A$: 1.5 Marks
• Factoring and cancelling bracket to get $\sec A + \tan A$: 1.0 Mark
• Converting to $1/(\sec A - \tan A)$: 1.0 Mark
Correct Answer: