Question

In $\tan \theta =\frac{40}{9}$, then $\sec \theta$ is :

• A. $\frac{41}{9}$
• B. $\frac{9}{41}$
• C. $\frac{9}{40}$
• D. None of these

Answer 1

Answer: (A)

$\frac{41}{9}$

We know that, $1+ta{n}^2\theta =se{c}^2\theta$

so, $sec\theta =\sqrt{1+ta{n}^2\theta }\Rightarrow \sqrt{1+\frac{1600}{81}}=\sqrt{\frac{1681}{81}}=\frac{41}{9}$