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Question: If \(\pi < \alpha < \frac{3\pi}{2}\), then \(\sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}} + \sqrt{\f...

If π<α<3π2\pi < \alpha < \frac{3\pi}{2}, then 1cosα1+cosα+1+cosα1cosα\sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}} + \sqrt{\frac{1 + \cos\alpha}{1 - \cos\alpha}}=

A

2sinα\frac{2}{\sin\alpha}

B

2sinα- \frac{2}{\sin\alpha}

C

1sinα\frac{1}{\sin\alpha}

D

1sinα- \frac{1}{\sin\alpha}

Answer

2sinα- \frac{2}{\sin\alpha}

Explanation

Solution

1cosα1+cosα+1+cosα1cosα=1cosα+1+cosα1cos2α\sqrt{\frac{1 - \cos\alpha}{1 + \cos\alpha}} + \sqrt{\frac{1 + \cos\alpha}{1 - \cos\alpha}} = \frac{1 - \cos\alpha + 1 + \cos\alpha}{\sqrt{1 - \cos^{2}\alpha}}

=2±sinα=2sinα,(since π<α<3π2).= \frac{2}{\pm \sin\alpha} = \frac{2}{- \sin\alpha},\left( \text{since }\pi < \alpha < \frac{3\pi}{2} \right).