Prove that $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 - \cos A}$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
LHS:
$$\frac{1 + \sec A}{\sec A} = \frac{1 + \dfrac{1}{\cos A}}{\dfrac{1}{\cos A}} = \frac{\dfrac{\cos A + 1}{\cos A}}{\dfrac{1}{\cos A}} = 1 + \cos A$$
RHS:
$$\frac{\sin^2 A}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A$$
Since LHS = RHS = $1 + \cos A$, the identity is proved. $\blacksquare$
Source: Exercise 8.3, Q4(iv), Chapter 8
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Explanation
- The hint in the textbook says "Simplify LHS and RHS separately" — examiners expect exactly this approach (not one-side manipulation).
- Key steps that earn marks: substituting $\sec A = \dfrac{1}{\cos A}$, and using the identity $\sin^2 A = 1 - \cos^2 A = (1-\cos A)(1+\cos A)$ on the RHS.
- Write LHS and RHS clearly labelled; state the conclusion "LHS = RHS" at the end — this is mandatory for full marks.