If sec theta + tan theta = frac{1}{3},then th | vedclass

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(C) Given $\sec 	heta + 	an 	heta = \frac{1}{3}$ . . . . . . $(i)$Since $\sec^2 	heta - 	an^2 	heta = 1$,we have $(\sec 	heta - 	an 	heta)(\s

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$3^{rd}$ quadrant

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(C) Given $\sec 	heta + 	an 	heta = \frac{1}{3}$ . . . . . . $(i)$Since $\sec^2 	heta - 	an^2 	heta = 1$,we have $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Thus,$\sec 	heta - 	an 	heta = 3$ . . . . . . $(ii)$Adding $(i)$ and $(ii)$,$2 \sec 	heta = \frac{1}{3} + 3 = \frac{10}{3} \Rightarrow \sec 	heta = \frac{5}{3}$.Subtracting $(ii)$ from $(i)$,$2 	an 	heta = \frac{1}{3} - 3 = \frac{-8}{3} \Rightarrow 	an 	heta = \frac{-4}{3}$.Since $\sec 	heta > 0$ and $	an 	heta Now,$	an 2 	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta} = \frac{2(-4/3)}{1 - 16/9} = \frac{-8/3}{-7/9} = \frac{24}{7} > 0$.Since $\sec 	heta = 5/3$,$\cos 	heta = 3/5$. Since $	heta$ is in the $4^{th}$ quadrant,$270^{\circ} Therefore,$540^{\circ} Since $	an 2 	heta > 0$ and $\sin 2 	heta = 2 \sin 	heta \cos 	heta = 2(-4/5)(3/5) = -24/25 < 0$,$2 	heta$ lies in the $3^{rd}$ quadrant (within the range $540^{\circ}$ to $720^{\circ}$).

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

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