Prove that :
$$\sqrt{\dfrac{\sec A - 1}{\sec A + 1}} + \sqrt{\dfrac{\sec A + 1}{\sec A - 1}} = 2\cosec A$$
Generated by claude-sonnet-4-6 · 2026-06-15 10:24 · grounding rag
Model Answer
LHS $= \sqrt{\dfrac{\sec A - 1}{\sec A + 1}} + \sqrt{\dfrac{\sec A + 1}{\sec A - 1}}$
$= \dfrac{(\sec A - 1) + (\sec A + 1)}{\sqrt{(\sec A+1)(\sec A-1)}}$
$= \dfrac{2\sec A}{\sqrt{\sec^2 A - 1}}$
Since $\sec^2 A - 1 = \tan^2 A$:
$= \dfrac{2\sec A}{\tan A} = \dfrac{2 \cdot \dfrac{1}{\cos A}}{\dfrac{\sin A}{\cos A}} = \dfrac{2}{\sin A} = 2\cosec A$ = RHS $\quad\blacksquare$
Source: Chapter 8, Section 8.4 Trigonometric Identities
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Explanation
- Key step: Combine the two surds over a common denominator $\sqrt{(\sec A+1)(\sec A-1)}$, giving numerator $(\sec A - 1)+(\sec A+1) = 2\sec A$.
- Identity used: $\sec^2 A - 1 = \tan^2 A$, so $\sqrt{\sec^2 A - 1} = \tan A$.
- Then simplify $\dfrac{\sec A}{\tan A} = \dfrac{1/\cos A}{\sin A/\cos A} = \dfrac{1}{\sin A} = \cosec A$.
- Examiners award marks for: correct combination of fractions, correct identity applied, and clean final simplification. Show each step clearly.