Q. 22

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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 1st quadrant

sin, cos & tan all are positive all are positive

Given

sin x = (1/4)

We know that

cos2x = 1 - sin2x

= 1 - (1/4)2

= 1 - (1/16)

cos2x = 15/16

cos x = (√15)/4

Since x is in IInd quadrant

cos x is negative

cos x = - (√15)/4

Also,

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

- (√15)/4 = 2 cos2(x/2) - 1

1- (√15)/4 = 2 cos2(x/2)

cos2(x/2) = [4 - (√15)]/8  lies in 1st quadrant is positive  Now,

We know that

1 + tan2x = sec2x

Replacing x with x/2       lies in 1st quadrant is positive in 1st quadrant We know that- Replacing x with x/2  Hence, Rate this question :

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