Question

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

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

Answer 1

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

$\frac{{\cos }^2\theta +{\tan }^2\theta -1}{{\sin }^2\theta }=\frac{co{s}^2\theta +ta{n}^2\theta -si{n}^2\theta -co{s}^2\theta }{si{n}^2\theta }=\frac{{\tan }^2\theta }{{\sin }^2\theta }-1=\frac{1}{co{s}^2\theta }-1=\frac{1-co{s}^2\theta }{co{s}^2\theta }=\frac{si{n}^2\theta }{co{s}^2\theta }={\tan }^2\theta$