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 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,



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