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

Question

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

Sapling Academy · Accepted Answer

Since, $\sec \alpha = \frac{2}{\sqrt3}$ $	herefore \cos \alpha = \frac{1}{\sec \alpha} =  \frac {1}{\frac{2}{\sqrt3}} = \frac{\sqrt 3}{2}$ We know that $\sin ^2 \alpha + \cos ^2 \alpha = 1$ $	herefore \sin ^2 \alpha = 1 -  \cos ^2 \alpha  = 1 - (\frac{\sqrt 3}{2})^2 = 1 - \frac{3}{4} = \frac{1}{4} $ $	herefore \sin \alpha = \pm \frac{1}{2}$ Now since, α is in IV quadrant, therefore value of $\sin \alpha$ will be negative only. $	herefore \sin \alpha =  - \frac{1}{2}$ $	herefore \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$ Please do press “Heart” button if you liked the solution.

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