Question

The expression $1+\frac{{\tan }^2\theta }{1+\sec \theta }$ when simplified reduces to

• A. $\sin \theta$
• B. $\sec \theta$
• C. $\text{cosec}\theta$
• D. $\cot \theta$

Answer 1

Answer: (B)

$\sec \theta$

$1+\frac{{\tan }^2\theta }{1+\sec \theta }$
= $1+\frac{{\sec }^2\theta -1}{1+\sec \theta }$ (Since, ${\sec }^2\theta -{\tan }^2\theta =1$)

= $1+\frac{(\sec \theta -1)(\sec \theta +1)}{1+\sec \theta }$

= $1+\sec \theta -1$
= $\sec \theta$