Chapter 14.2, Problem 3CYP

To determine

To Verify :The given equationYY ${\mathrm{cosec}}^2\theta -{\cot }^2\theta =\cot \theta \tan \theta$

Explanation of Solution

Given information :

We know that, the given value is ${\mathrm{cosec}}^2\theta -{\cot }^2\theta =\cot \theta \tan \theta$

Formula Used:

  $1-{\cos }^2\theta ={\sin }^2\theta$

Calculation :

The given equation is

  $\begin{array}{l}L.H.S.={\mathrm{cosec}}^2\theta -{\cot }^2\theta \\ =\frac{1}{{\sin }^2\theta }-\frac{{\cos }^2\theta }{{\sin }^2\theta }\\ =\frac{1-{\cos }^2\theta }{{\sin }^2\theta }=\frac{{\sin }^2\theta }{{\sin }^2\theta }=1\\ R.H.S.=\cot \theta \tan \theta \\ =\frac{1}{\tan \theta }\times \tan \theta \\ =1\\ \text{So, }{\mathrm{cosec}}^2\theta -{\cot }^2\theta =\cot \theta \tan \theta \end{array}$