Chapter 14.1, Problem 48HP

To determine

To Prove:Rewrite ${\text{sin}}^2{\text{θ+cos}}^2\text{θ=1}\text{as}1+{\text{cot}}^2\text{θ}={\text{cosec}}^2\text{θ}$

Explanation of Solution

Given

We know that, the given expression is ${\text{sin}}^2{\text{θ+cos}}^2\text{θ=1}\text{as}1+{\text{cot}}^2\text{θ}={\text{cosec}}^2\text{θ}$

Calculation :

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

Above equation divided by ${\sin }^2\theta$ in both side.

  $\begin{array}{l}\frac{\cancel{{\sin }^2\theta }}{\cancel{{\sin }^2\theta }}+\frac{{\cos }^2\theta }{{\sin }^2\theta }=\frac{1}{{\sin }^2\theta }\\ 1+{\left(\frac{\cos \theta }{\cos \theta }\right)}^2={\left(\frac{1}{\sin \theta }\right)}^2\\ 1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta \end{array}$

Hence proved.