If sec theta + tan theta = 2/3,then in which | vedclass

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(D) Given that $\sec 	heta + 	an 	heta = 2/3$  $(i)$ We know that $\sec^2 	heta - 	an^2 	heta = 1$,which implies $(\sec 	heta - 	an 	heta)(\s

Vedclass · Accepted Answer

$IV$

Vedclass · Answer

(D) Given that $\sec 	heta + 	an 	heta = 2/3$  $(i)$ We know that $\sec^2 	heta - 	an^2 	heta = 1$,which implies $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$. Substituting the value from $(i)$,we get $\sec 	heta - 	an 	heta = 3/2$  (ii) Adding $(i)$ and (ii): $2 \sec 	heta = 2/3 + 3/2 = (4 + 9)/6 = 13/6$,so $\sec 	heta = 13/12$. Subtracting (ii) from $(i)$: $2 	an 	heta = 2/3 - 3/2 = (4 - 9)/6 = -5/6$,so $	an 	heta = -5/12$. Since $\sec 	heta > 0$ and $	an 	heta < 0$,$	heta$ must lie in the $IV$ quadrant.

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

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