Question

IfTantheta+ Sectheta
√3, then theta=​

Answer 1

Given,

$\longrightarrow\sec\theta+\tan\theta=\sqrt3\quad\quad\dots(1)$

We know that,

$\longrightarrow \sec^2\theta-\tan^2\theta=1$

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

From (1),

$\longrightarrow\sqrt3\,(\sec\theta-\tan\theta)=1$

$\longrightarrow\sec\theta-\tan\theta=\dfrac{1}{\sqrt3}\quad\quad\dots(2)$

Adding (1) and (2),

$\longrightarrow(\sec\theta+\tan\theta)+(\sec\theta-\tan\theta)=\sqrt3+\dfrac{1}{\sqrt3}$

$\longrightarrow2\sec\theta=\dfrac{4}{\sqrt3}$

$\longrightarrow\sec\theta=\dfrac{2}{\sqrt3}$

$\longrightarrow\sec\theta=\sec\left(\dfrac{\pi}{6}\right)$

Subtracting (2) from (1),

$\longrightarrow(\sec\theta+\tan\theta)-(\sec\theta-\tan\theta)=\sqrt3-\dfrac{1}{\sqrt3}$

$\longrightarrow2\tan\theta=\dfrac{2}{\sqrt3}$

$\longrightarrow\tan\theta=\dfrac{1}{\sqrt3}$

$\longrightarrow\tan\theta=\tan\left(\dfrac{\pi}{6}\right)$

We get $\theta$ lies in first quadrant since $\sec\theta$ and $\tan\theta$ are positive.

So general value of $\theta$ is,

$\longrightarrow\underline{\underline{\theta=2n\pi+\dfrac{\pi}{6}}},\quad n\in\mathbb{Z}$

And principal value of $\theta$ is,

$\longrightarrow\underline{\underline{\theta=\dfrac{\pi}{6}}}$