Q60 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics
If $\sec \theta+\tan \theta=\frac{1}{3}$, then the quadrant in which $2 \theta$ lies is
- A$1^{\text {st }}$ quadrant
- B$2^{\text {nd }}$ quadrant
- Correct answerC$3^{\text {rd }}$ quadrant
- D$4^{\text {th }}$ quadrant
Solution
$\sec \theta+\tan \theta=\frac{1}{3}$ $\qquad ....\mathrm{(i)}$
$\frac{1}{\sec \theta-\tan \theta}=\frac{1}{3} \Rightarrow \sec \theta-\tan \theta=3$ $\qquad ....\mathrm{(ii)}$
From (i) - (ii) $\begin{aligned} & 2 \tan \theta=\frac{1}{3}-3=\frac{-8}{3} \Rightarrow \tan \theta=\frac{-4}{3} \\ & \tan 2 \theta=\frac{\frac{-8}{3}}{1-\frac{16}{9}}=\frac{24}{7}\gt0 \end{aligned}$ $2 \theta$ lies in $3^{\text {rd }}$ quadrants
From (i) - (ii) $\begin{aligned} & 2 \tan \theta=\frac{1}{3}-3=\frac{-8}{3} \Rightarrow \tan \theta=\frac{-4}{3} \\ & \tan 2 \theta=\frac{\frac{-8}{3}}{1-\frac{16}{9}}=\frac{24}{7}\gt0 \end{aligned}$ $2 \theta$ lies in $3^{\text {rd }}$ quadrants
Hence, the correct answer is option: 'C'.