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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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?

Similar Questions

$\sin \frac{\pi}{16} \sin \frac{3 \pi}{16} \sin \frac{5 \pi}{16} \sin \frac{7 \pi}{16}$ is equal to

The number of all possible integral values of $n > 2$ such that $\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$ is:

$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = $

In a right-angled triangle,the hypotenuse is $2 \sqrt{2}$ times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are

If $\cos x + \sin x = \frac{1}{2}$ and $0 < x < \pi$,then $\tan x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module