If sec α = 2/√3, then find the value of 1- cosec α /1+ cosec α , where α is in IV quadrant.

If sec α = (2 )/√3, then find the value of 1- cosec α /1+ cosec α

Q) If $\sec \alpha = \frac{2}{\sqrt3}$ , then find the value of $\frac {1- \csc \alpha}{1+ \csc \alpha}$ , where α is in IV quadrant.

Ans:

Since, $\sec \alpha = \frac{2}{\sqrt3}$

$\therefore \cos \alpha = \frac{1}{\sec \alpha} = \frac {1}{\frac{2}{\sqrt3}} = \frac{\sqrt 3}{2}$

We know that $\sin ^2 \alpha + \cos ^2 \alpha = 1$

$\therefore \sin ^2 \alpha = 1 - \cos ^2 \alpha = 1 - (\frac{\sqrt 3}{2})^2 = 1 - \frac{3}{4} = \frac{1}{4}$

$\therefore \sin \alpha = \pm \frac{1}{2}$

Now since, α is in IV quadrant, therefore value of $\sin \alpha$ will be negative only.

$\therefore \sin \alpha = - \frac{1}{2}$

$\therefore \csc \alpha = - 2$

Now, $\frac {1 - \csc \alpha}{1 + \csc \alpha} = \frac {1 - (- 2)}{1 + (- 2)} = \frac {1 + 2}{1 - 2} = \frac {3}{-1} = -3$

Therefore ${\frac {1- \csc \alpha}{1+ \csc \alpha} = -3$

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