303
If $\sin x = p$, then prove that :
(i) $\cot x = \frac{\sqrt{1-p^2}}{p}$
(ii) $\frac{1 + \tan^2 x}{1+\cot^2 x} = \frac{p^2}{1-p^2}$
(i) $\cot x = \frac{\sqrt{1-p^2}}{p}$
(ii) $\frac{1 + \tan^2 x}{1+\cot^2 x} = \frac{p^2}{1-p^2}$
Show SolutionHide Solution↓
(i) $\cot x = \frac{\cos x}{\sin x} = \frac{\sqrt{1 - \sin^2 x}}{\sin x}$ (1 Mark)
$= \frac{\sqrt{1-p^2}}{p}$ (1/2 Mark)
(ii) $\frac{1+\tan^2 x}{1+\cot^2 x} = \frac{\sec^2 x}{\operatorname{cosec}^2 x} = \frac{\sin^2 x}{\cos^2 x}$ (1/2 + 1/2 Mark)
$= \frac{p^2}{1-p^2}$ (1/2 Mark)
$= \frac{\sqrt{1-p^2}}{p}$ (1/2 Mark)
(ii) $\frac{1+\tan^2 x}{1+\cot^2 x} = \frac{\sec^2 x}{\operatorname{cosec}^2 x} = \frac{\sin^2 x}{\cos^2 x}$ (1/2 + 1/2 Mark)
$= \frac{p^2}{1-p^2}$ (1/2 Mark)