For any trigonometric point P(x,y)
x always represents cos
[tex] x=cos\theta [/tex]
y always represents sin.
[tex] y=sin\theta [/tex]
Now if we drop a perpendicular from P(x,y) to a point Q which is a refelction of P across x axis, we get Q(x, -y) for the same angle.
The angle shall be [tex] -\theta [/tex]
So now
[tex] cos(\-theta)=x $ and $ sin(-\theta) = -y [/tex] ...(1)
But [tex] x=cos\theta $ and $ y = sin\theta [/tex]
Statement (1) becomes
[tex] cos(-\theta) = cos\theta $ and $ sin(-\theta) = -sin\theta [/tex]
So the value of cos does not change, but the value of sin changes.
Cos is even & sin is odd.
And so sec is even and cosec is odd.
So [tex] f(x) = csc(\frac{-3\pi}{2}) [/tex] shall be an odd function.
Option B) is the right answer
