Question

$ta{n}^2\theta -se{c}^2\theta$ is equal to

• A. 1
• B. -1
• C. 0
• D. $co{t}^2\theta$

Answer 1

Answer: (B)

-1

As we know that,

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

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

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

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