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If $\sec \theta+\tan \theta=2 / 3$, then in which quadrant does $\theta$ lie in?
Options:
Solution:
1730 Upvotes
Verified Answer
The correct answer is:
IV
$$
\begin{aligned}
& \text { } \sec \theta+\tan \theta=\frac{2}{3} \\
& \Rightarrow \sec \theta-\tan \theta=\frac{3}{2}
\end{aligned}
$$
Eqs. (i) + (ii)
$$
\Rightarrow \quad 2 \sec \theta=\frac{13}{6} \Rightarrow \sec \theta=\frac{13}{12}
$$
Eqs. (i) -(ii)
$$
\Rightarrow \quad 2 \tan \theta=-\frac{5}{6} \Rightarrow \tan \theta=-\frac{5}{12}
$$
$\therefore \sec \theta$ is positive and $\tan \theta$ is negative
$\therefore \theta$ lies in IV quadrant
Hence, option (4) is correct.
\begin{aligned}
& \text { } \sec \theta+\tan \theta=\frac{2}{3} \\
& \Rightarrow \sec \theta-\tan \theta=\frac{3}{2}
\end{aligned}
$$
Eqs. (i) + (ii)
$$
\Rightarrow \quad 2 \sec \theta=\frac{13}{6} \Rightarrow \sec \theta=\frac{13}{12}
$$
Eqs. (i) -(ii)
$$
\Rightarrow \quad 2 \tan \theta=-\frac{5}{6} \Rightarrow \tan \theta=-\frac{5}{12}
$$
$\therefore \sec \theta$ is positive and $\tan \theta$ is negative
$\therefore \theta$ lies in IV quadrant
Hence, option (4) is correct.
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