Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
\((cosec θ – cot θ)^2 = \frac{1 – cosθ}{1 + cosθ}\)
Explanation:
We know that, \(cosec \theta=\frac{1}{sin \theta}\) and \(cot \theta=\frac{cos \theta}{sin \theta}\)
Therefore, \((cosec θ – cot θ)^2\)=
[\(sin^2 \theta=1-cos^2 \theta\)]
Therefore, \((cosec \theta – cot \theta)^2\) =
\(= \frac{1-cos \theta}{1+cos \theta}\)
\(=\) RHS , Hence Proved.
Answer variants:
\(\frac{(1-cos\theta)^2}{(1-cos \theta)(1+cos \theta)}\)
\(= \frac{(1-cos \theta)^2}{sin^2\theta}\)
\(= \frac{(1-cos \theta)^2}{(1-cos \theta)^2}\)