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Q) Prove that: \frac{sec^3 \theta}{(\sec^2 \theta - 1)} + \frac{cosec^3 \theta}{(\cos ec^2 \theta - 1)} = sec θ . cosec θ (sec θ + cosec θ)

(Q 26 A – 30/2/2 – CBSE 2026 Question Paper)

Ans: 

Step 1: Let’s start from LHS:

\frac{sec^3 \theta}{(\sec^2 \theta - 1)} + \frac{cosec^3 \theta}{(\cos ec^2 \theta - 1)}

= \frac{\frac{1}{\cos ^3 \theta}} {(\frac{1}{\cos ^2 \theta} - 1)} + \frac{\frac{1}{\sin ^3 \theta}} {(\frac{1}{\sin ^2 \theta} - 1)}

= \frac{\frac{1}{\cos ^3 \theta}} {(\frac{(1 - \cos ^2 \theta)}{\cos ^2 \theta})} + \frac{\frac{1}{\sin ^3 \theta}} {(\frac{(1 - \sin ^2 \theta)}{\sin ^2 \theta})}

= \frac{\cos ^2 \theta}{\cos ^3 \theta (1 - \cos ^2 \theta)} + \frac{\sin ^2 \theta}{\sin ^3 \theta (1 - \sin ^2 \theta)}

= \frac{1}{\cos \theta (1 - \cos ^2 \theta)} + \frac{1}{\sin \theta (1 - \sin ^2 \theta)}

Step 2: Since sin 2 θ + cos 2 θ = 1

∴ 1 – cos 2 θ = sin 2 θ and 1 – sin 2 θ = cos 2 θ

Step 3: By substituting these values, we get:

LHS = \frac{1}{\cos \theta (\sin ^2 \theta)} + \frac{1}{\sin \theta (\cos ^2 \theta)}

= \frac{1}{\sin \theta \cos \theta} (\frac{1}{\sin \theta} + \frac{1}{\cos \theta})

= cosec θ . sec θ (cosec θ + sec θ)

= sec θ . cosec θ (sec θ + cosec θ) = RHS

Hence Proved!

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