cos x = –1/3, x in quadrant III
Given that x is in quadrant II
So,
180° < x < 270°
Dividing with 2 all sides
(180°/2) < x/2 < (270°/2)
90° < x/2 < 135°
∴ lies in 2nd quadrant
In 2nd quadrant,
sin is positive and cos, tan are negative
⇒ is positive and are negative
Given
cos x = -(1/3)
We know that
cos 2x = 2 cos2x - 1
Replacing x with x/2
cos 2(x/2) = 2 cos2(x/2) - 1
⇒ cos x = 2 cos2(x/2) - 1
⇒ -1/3 = 2 cos2(x/2) - 1
⇒ 2/3 = 2 cos2(x/2)
⇒ cos2(x/2) = 1/3
∴ cos(x/2) = (1/√3)
∵ lies in 2nd quadrant
is negative
∴
Now,
We know that
1 + tan2x = sec2x
Replacing x with x/2
∵ lies in 2nd quadrant
is positive in 2nd quadrant
We know that-
Replacing x with x/2
Hence,