Answer :

Given that x is in quadrant II

So,

90° < x < 180°

Dividing with 2 all sides

(90°/2) < x/2 < (180°/2)

45° < x/2 < 90°

∴ lies in 1^{st} quadrant

In 1^{st} quadrant,

sin, cos & tan all are positive

⇒ all are positive

Given

sin x = (1/4)

We know that

cos^{2}x = 1 - sin^{2}x

= 1 - (1/4)^{2}

= 1 - (1/16)

∴ cos^{2}x = 15/16

⇒ cos x = (√15)/4

Since x is in II^{nd} quadrant

∴ cos x is negative

⇒ cos x = - (√15)/4

Also,

cos 2x = 2 cos^{2}x - 1

Replacing x with x/2

cos 2(x/2) = 2 cos^{2}(x/2) - 1

⇒ cos x = 2 cos^{2}(x/2) - 1

⇒ - (√15)/4 = 2 cos^{2}(x/2) - 1

⇒ 1- (√15)/4 = 2 cos^{2}(x/2)

⇒ cos^{2}(x/2) = [4 - (√15)]/8

∵ lies in 1^{st} quadrant

is positive

Now,

We know that

1 + tan^{2}x = sec^{2}x

Replacing x with x/2

∵ lies in 1^{st} quadrant

is positive in 1^{st} quadrant

We know that-

Replacing x with x/2

Hence,

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