Question

The value of $\frac{{\cos }^2\theta +{\tan }^2\theta -1}{{\sin }^2\theta }$ is

• A. $0$
• B. ${\cos }^2\theta$
• C. ${\tan }^2\theta$
• D. $\frac{1}{{\sin }^2\theta }$

Answer 1

The correct option is C ${\tan }^2\theta$

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

$={\cot }^2\theta +{\text{sec}}^2\theta -{\text{cosec}}^2\theta$

$={\text{sec}}^2\theta -({\text{cosec}}^2\theta -{\cot }^2\theta )$

$={\text{sec}}^2\theta -1$

$={\tan }^2\theta$