Question

Number of solution of $\sec \theta +\tan \theta =2+\sqrt{3}$ in the interval $(0,2\pi )$ is

• A. $1$
• B. $0$
• C. $2$
• D. $noneofthese$

Answer 1

Answer: (A)

$1$

$\sec \theta =\tan \theta =2+\sqrt{3}⟶\left(1\right)$

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

$\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$

$(\sec \theta -\tan \theta )(2+\sqrt{3})=1$

$\sec \theta -\tan \theta =\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

$\sec \theta -\tan \theta =\frac{2-\sqrt{3}}{4-{\left(\sqrt{3}\right)}^2}=2-\sqrt{3}$

$\sec \theta -\tan \theta =2-\sqrt{3}⟶\left(II\right)$

Solving $\left(I\right)\&\left(II\right)$

$\Rightarrow 2\sec \theta =4$

$\sec \theta =2$

and $2\tan \theta =2\sqrt{3}$

$\tan \theta =\sqrt{3}$

As interval is $0$ to $2r$

Both $\sec \theta \&\tan \theta$ can be positive in first quadrant only.

So, only one value.

Hence, the answer is $1.$