Question

State true or false.

$\frac{{\tan }^2\theta }{{\sec }^2\theta }+\frac{{\cot }^2\theta }{cose{c}^2\theta }=1$

• A. True
• B. False

Answer 1

The correct option is A True

Given,

$\frac{ta{n}^2\theta }{se{c}^2\theta }+\frac{co{t}^2\theta }{cs{c}^2\theta }$

$=\frac{{\sin }^2\theta }{{\cos }^2\theta }\times {\cos }^2\theta +\frac{{\cos }^2\theta }{{\sin }^2\theta }\times {\sin }^2\theta$

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

$=1$