Prove the following trigonometry identity (sin θ + cos θ)

Q) Prove the following trigonometry identity: (sin θ + cos θ)(cosec θ – sec θ) = cosec θ. sec θ – 2 tan θ

ICSE Specimen Question Paper (SQP)2025

Ans:

Let’s start from LHS:

(sin θ + cos θ)(cosec θ – sec θ)

= sin θ cosec θ – sin θ sec θ + cos θ cosec θ – cos θ sec θ

= 1 – $\frac{\sin \theta}{\cos \theta} + \frac {\cos \theta}{\sin \theta}$ – 1

= $\frac{\cos ^2 \theta - \sin ^2 \theta}{\sin \theta . \cos \theta}$

We know that sin ² θ + cos ² θ = 1

∴ cos ²θ = 1 – sin ²θ

by substituting the above in above equation, we get:

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

= $\frac{1 - 2 \sin ^2 \theta}{\sin \theta . \cos \theta}$

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

= $\frac{1}{\sin \theta} . \frac{1}{\cos \theta} - 2 (\frac{\sin \theta}{\cos \theta})$

= cosec θ sec θ – 2 tan θ

= sec θ cosec θ – 2 tan θ

= RHS ………. hence proved !

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