Chapter 14, Problem 18MCQ

To determine

To verify: The given identity.

Explanation of Solution

Given:

The identity is given as:

  ${\sin }^2\theta .{\tan }^2\theta ={\tan }^2\theta -{\sin }^2\theta$

Consider the identify:

  ${\sin }^2\theta .{\tan }^2\theta ={\tan }^2\theta -{\sin }^2\theta$

From identify:

  $\tan \theta =\frac{\sin \theta }{\cos \theta }$

Take the left hand side of the identify:

  $\begin{array}{l}{\sin }^2\theta .{\tan }^2\theta ={\sin }^2\theta .\frac{{\sin }^2\theta }{{\cos }^2\theta }\\ =\left({\sin }^2\theta \right)\frac{1}{{\cos }^2\theta }\end{array}$

From identify:

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

Therefore,

  $\begin{array}{l}{\sin }^2\theta .{\tan }^2\theta ={\sin }^2\theta .\frac{{\sin }^2\theta }{{\cos }^2\theta }\\ =\left({\sin }^2\theta \right)\frac{1-{\cos }^2\theta }{{\cos }^2\theta }\\ ={\sin }^2\theta \left(\frac{1}{{\cos }^2\theta }-1\right)\\ =\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta \end{array}$

Solving further,

  $\begin{array}{l}{\sin }^2\theta .{\tan }^2\theta ={\left(\frac{\sin \theta }{\cos \theta }\right)}^2-{\sin }^2\theta \\ {\tan }^2\theta -{\sin }^2\theta \\ =\text{Right}\text{Hand}\text{Side}\end{array}$

Hence, from the above equation, the above identify is verified.