If $\sec heta+ an heta=\frac{1}{3}$, then the quadrant in which $2 heta$ lies is

Question

If $\sec 	heta+	an 	heta=\frac{1}{3}$, then the quadrant in which $2 	heta$ lies is

Aicharya · Accepted Answer

$\sec 	heta+	an 	heta=\frac{1}{3}$ $\qquad ....\mathrm{(i)}$ $\frac{1}{\sec 	heta-	an 	heta}=\frac{1}{3} \Rightarrow \sec 	heta-	an 	heta=3$ $\qquad ....\mathrm{(ii)}$ From (i) - (ii) $\begin{aligned} & 2 	an 	heta=\frac{1}{3}-3=\frac{-8}{3} \Rightarrow 	an 	heta=\frac{-4}{3} \ & 	an 2 	heta=\frac{\frac{-8}{3}}{1-\frac{16}{9}}=\frac{24}{7}\gt0 \end{aligned}$ $2 	heta$ lies in $3^{	ext {rd }}$ quadrants

If $\sec \theta+\tan \theta=\frac{1}{3}$, then the quadrant in which $2 \theta$ lies is

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