183
If $\text{cosec } \theta = x + \frac{1}{4x}$, prove that $\text{cosec } \theta + \cot \theta = 2x$ or $\frac{1}{2x}$.
Show SolutionHide Solution↓
$\cot^2 \theta = \text{cosec}^2 \theta - 1 = (x + \frac{1}{4x})^2 - 1$ ($1$)
$= (x - \frac{1}{4x})^2$ ($1/2$)
$\Rightarrow \cot \theta = (x - \frac{1}{4x})$ or $(-x + \frac{1}{4x})$ ($1/2$)
$\text{cosec } \theta + \cot \theta = (x + \frac{1}{4x}) + (x - \frac{1}{4x})$ or $(x + \frac{1}{4x}) + (-x + \frac{1}{4x})$ ($1/2+1/2$)
$= 2x$ or $\frac{1}{2x}$
$= (x - \frac{1}{4x})^2$ ($1/2$)
$\Rightarrow \cot \theta = (x - \frac{1}{4x})$ or $(-x + \frac{1}{4x})$ ($1/2$)
$\text{cosec } \theta + \cot \theta = (x + \frac{1}{4x}) + (x - \frac{1}{4x})$ or $(x + \frac{1}{4x}) + (-x + \frac{1}{4x})$ ($1/2+1/2$)
$= 2x$ or $\frac{1}{2x}$