Question

1/cosec theta + cot theta -1/ sin theta=1/sin theta - 1/ cosec theta -cot
theta
​

Answer 1

`\theta =2n\pi, \;\frac{\pi}2+2n\pi \;\;\text{for all}\; n=0,1,2,...`

Explanation:

`\frac 1{\text{cosec} \theta}+\cot \theta-\frac 1{\sin \theta}=\frac 1{\sin \theta}-\frac 1{\text{cosec} \theta}-\cot \theta`

`\Rightarrow \frac 2{\text{cosec} \theta}+2\cot \theta-\frac 2{\sin \theta}=0`

`\Rightarrow \frac 1{\text{cosec} \theta}+\cot \theta-\frac 1{\sin \theta}=0`

`\Rightarrow \sin \theta -\frac 1{\sin \theta}+(\cos \theta)/(\sin \theta)=0`

`\Rightarrow (\sin ^2\theta-1+\cos \theta)/(\sin \theta)=0`

`\Rightarrow (-\cos ^2\theta+\cos \theta)/(\sin \theta)=0\Rightarrow(\cos \theta-\cos ^2\theta)/(\sin \theta)=0`

`\Rightarrow \cot \theta(1-\cos \theta)=0`

`\text{If} \;\;1-\cos \theta=0\Rightarrow \cos \theta=1`

`\Rightarrow \theta=2n\pi\;\; \text{for all}\; n=0,1,2,.....`

`\text{If} \cot \theta=0\Rightarrow (\cos \theta)/(\sin \theta)=0\Rightarrow \cos \theta=0`

`\Righarrow \theta =\frac{\pi}2+2n\pi\;\;\text{for all}\; n=0,1,2,...`

Hence,
`\theta =2n\pi, \;\frac{\pi}2+2n\pi \;\;\text{for all}\; n=0,1,2,...`