If $\sec heta+ an heta=2 / 3$, then in which quadrant does $ heta$ lie in?

Question

If $\sec 	heta+	an 	heta=2 / 3$, then in which quadrant does $	heta$ lie in?

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } \sec 	heta+	an 	heta=\frac{2}{3} \ & \Rightarrow \sec 	heta-	an 	heta=\frac{3}{2} \end{aligned} $ Eqs. (i) + (ii) $ \Rightarrow \quad 2 \sec 	heta=\frac{13}{6} \Rightarrow \sec 	heta=\frac{13}{12} $ Eqs. (i) -(ii) $ \Rightarrow \quad 2 	an 	heta=-\frac{5}{6} \Rightarrow 	an 	heta=-\frac{5}{12} $ $	herefore \sec 	heta$ is positive and $	an 	heta$ is negative $	herefore 	heta$ lies in IV quadrant Hence, option (4) is correct

If $\sec \theta+\tan \theta=2 / 3$, then in which quadrant does $\theta$ lie in?

Solution