Question

lf $\alpha$ lies in the third quadrant, then $\sqrt{\frac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }}=$

• A. $-2\tan \alpha$
• B. $-2\sec \alpha$
• C. $2\cot \alpha$
• D. $2\tan \alpha$

Answer 1

The correct option is B $-2\sec \alpha$

$\sqrt{\frac{1-\sin \alpha }{1+\sin \alpha }}+\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }}$

$=\sqrt{\frac{1-\sin \alpha }{1+\sin \alpha }\times \frac{1-\sin \alpha }{1-\sin \alpha }}+\sqrt{\frac{1+\sin \alpha }{1-\sin \alpha }\times \frac{1+\sin \alpha }{1+\sin \alpha }}$

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

$=\left|\frac{(1-\sin \alpha )}{\cos \alpha }\right|+\left|\frac{(1+\sin \alpha )}{\cos \alpha }\right|$

$=|\sec \alpha -\tan \alpha |+|\sec \alpha +\tan \alpha |$

Since $\alpha$ lies in the third quadrant, both $\sec \alpha$ and $\tan \alpha$ will be negative, and also $\tan \alpha >\sec \alpha$

Therefore, removing the modulus gives,

$\tan \alpha -\sec \alpha -\sec \alpha -\tan \alpha =-2\sec \alpha$