Find in the following:

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 2^nd quadrant

In 2^nd 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 2^nd quadrant

is negative

∴

Now,

We know that

1 + tan²x = sec²x

Replacing x with x/2

∵ lies in 2^nd quadrant

is positive in 2^nd quadrant

We know that-

Replacing x with x/2

Hence,