Question

$(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )=$

• A. ${\cos }^2\theta -{\sin }^2\theta$
• B. ${\sin }^2\theta -{\cos }^2\theta$
• C. ${\sin }^2\theta +{\cos }^2\theta$
• D. 0

Answer 1

The correct option is C ${\sin }^2\theta +{\cos }^2\theta$

We have,

$(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )$

$\Rightarrow (1+{\tan }^2\theta )(1-{\sin }^2\theta )$ $\because a^2-b^2=(a+b)(a-b)$

$\Rightarrow (1+{\tan }^2\theta )\left({\cos }^2\theta \right)$ $\because {\cos }^2\theta =1-{\sin }^2\theta$

$\Rightarrow \left({\sec }^2\theta \right)\left({\cos }^2\theta \right)$ $\because 1+{\tan }^2\theta =se{c}^2\theta$

$\Rightarrow {\sec }^2\theta \times \frac{1}{{\sec }^2\theta }$ $\because {\cos }^2\theta =\frac{1}{{\sec }^2\theta }$

$\Rightarrow 1$

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

Hence, this is the answer.