Question

Prove the following trigonometric identities:

`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `

Answer 1

(i)  We have,

`LHS=\frac{\sin \theta }{(1-\cos \theta )}\times \frac{(1+\cos \theta)}{(1+\cos \theta )}`

[Multiplying numerator and denominator by (1 + cosθ)]

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

[∵ 1 – cos² θ = sin² θ]

`=\frac{1+\cos \theta }{\sin \theta }=\frac{1}{\sin \theta }+\frac{\cos\theta }{\sin \theta }`

= cosecθ + cotθ = RHS

`[ \because \ \ \frac{1}{\sin \theta }=\cos ec\theta `

(ii) We have,

`LHS=\frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta }`

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

`\frac{\frac{1}{\cos \theta }+1}{\frac{1}{\cos \theta }-1}=\frac{\sec\theta +1}{\sec \theta -1}=RHS`