Question

$(1+{\cot }^2\theta )(1-\cos \theta )(1+\cos \theta )=$

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

Answer 1

Answer: (C)

${\sec }^2\theta -{\tan }^2\theta$

We have,

$(1+{\cot }^2\theta )(1-\cos \theta )(1+\cos \theta )$

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

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

$\Rightarrow \left({\csc }^2\theta \right)\left({\sin }^2\theta \right)$ $\because 1+{\cot }^2\theta ={\csc }^2\theta$

$\Rightarrow \frac{1}{{\sin }^2\theta }\times {\sin }^2\theta$ $\because {\csc }^2\theta =\frac{1}{{\sin }^2\theta }$

$\Rightarrow 1$

$\Rightarrow {\sec }^2\theta -{\tan }^2\theta$ $\because {\sec }^2\theta -{\tan }^2\theta =1$

Hence, this is the answer.