Question

${\sin }^2\theta +\frac{1}{1+{\tan }^2\theta }=$

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

Answer 1

The correct option is B $cose{c}^2\theta -{\cot }^2\theta$

We have,

${\sin }^2\theta +\frac{1}{1+{\tan }^2\theta }$

$\Rightarrow {\sin }^2\theta +\frac{1}{{\sec }^2\theta }$ $\because 1+{\tan }^2\theta ={\sec }^2\theta$

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

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

$\Rightarrow {\csc }^2\theta -{\cot }^2\theta$ $\because {\csc }^2\theta -{\cot }^2\theta =1$

Hence, this is the answer.