Question

Find the value of
$\quad \displaystyle\frac{1}{1-\sin\theta}+\displaystyle\frac{1}{1+\sin\theta} = 2\sec^2\theta$

• A. $2{\sec }^2\theta$
• B. $2{\tan }^2\theta$
• C. $2{\cos }^2\theta$
• D. $2cose{c}^2\theta$

Answer 1

The correct option is A $2{\sec }^2\theta$

Simplifying the given expression, we get
$\frac{1+sin\theta +1-sin\theta }{1-si{n}^2\theta }$
$=\frac{2}{co{s}^2\theta }$
$=2se{c}^2\theta$