26 a. Prove that: sec^3 θ / (sec^2 θ – 1) + cosec^3 θ / (cosec^2 θ – 1) = sec θ . cosec θ (sec θ + cosec θ)

Question

Q) Prove that: $\frac{sec^3 	heta}{(\sec^2 	heta - 1)} + \frac{cosec^3 	heta}{(\cos ec^2 	heta - 1)}$ = sec θ . cosec θ (sec θ + cosec θ)
(Q 26 A - 30/2/2 - CBSE 2026 Question Paper)

Sapling Academy · Accepted Answer

Step 1: Let's start from LHS:

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

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

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

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

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

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 = $\frac{1}{\cos 	heta (\sin ^2 	heta)} + \frac{1}{\sin 	heta (\cos ^2 	heta)}$

= $\frac{1}{\sin 	heta \cos 	heta} (\frac{1}{\sin 	heta} + \frac{1}{\cos 	heta})$

= cosec θ . sec θ (cosec θ + sec θ)

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

Hence Proved!

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