Respuesta :
Here sin theta is positive and tan theta is negative , it means theta is in second quadrant where cos is negative too .
[tex] sin \Theta = 2/3 , cos \theta = - \sqrt{1-sin^2 \Theta} =- \sqrt{1-4/9}=-\sqrt{5}/3 [/tex]
So the correct option is the third option .
It is given tan theta <0
we know that tan theta is negative ( OR <0 ) in second ( pi/2 to pi )and fourth quadrant ( from 3pi/2 to 2 pi ) only .
Given sin theta = 2/3 which is positive ( first and second quadrant ) value so we can consider second quadrant only
Second quadrant tan( theta ) is <0 and sin( theta ) >0
We know sin ^2 ( theta) + cos^2( theta) = 1
( 2/3) ^2 + cos^2( theta ) = 1
Cos^2 (theta ) = 1- ( 2/3)^2
Cos^2 ( theta ) = 1- (4/9)
Cos^2 (theta ) = ( 9-4) / 9
Cos^2 theta = 5 /9
Cos theta = sqrt 5/3 OR - sqrt 5 /3
BUT Cos is negative in second quadrant
So answer is - sqrt 5/ 3