Q) Prove that: √(1 – sin θ)/(1 + sin θ) = sec θ – tan θ. (Q 24 B – 30/2/1 – CBSE 2026 Question Paper) Ans: Let’s start from LHS: LHS = Step 1: [Note: Since in RHS, we need to get cos θ in denominator, hence, we need to nullify + sign] ∴ […]

Prove that: √(1 - sin θ)/(1 + sin θ) = sec θ

Q) Prove that: √(1 – sin θ)/(1 + sin θ) = sec θ – tan θ.

(Q 24 B – 30/2/1 – CBSE 2026 Question Paper)

Ans:

Let’s start from LHS:

LHS = $\sqrt\frac{(1 - \sin \theta)}{(1 + \sin \theta)}$

Step 1: [Note: Since in RHS, we need to get cos θ in denominator, hence, we need to nullify + sign]

∴ Let’s both multiply & divide the expression by (1 – sin θ)

LHS = $\sqrt\frac{(1 - \sin \theta)}{(1 + \sin \theta)}$

∴ LHS = $\sqrt\frac{(1 - \sin \theta)(1 - \sin \theta)}{(1 + \sin \theta)(1 - \sin \theta)}$

= $\sqrt\frac{(1 - \sin \theta)^2}{(1 - \sin^2 \theta)}$ (∵ (a + b)(a – b) = a ² + b ²)

Step 2: ∵ we know that sin ² θ + cos ² θ = 1

∴ cos ² θ = 1 – sin ² θ

∴ LHS = $\sqrt\frac{(1 - \sin \theta)^2}{(\cos^2 \theta)}$

= $\frac{(1 - \sin \theta)}{(\cos \theta)}$

= $\frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}$

= sec θ – tan θ = RHS

Hence Proved !

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