MCQ
If $\alpha$ lies in the second quadrant,then $\sqrt {\frac{{1 - \sin \alpha }}{{1 + \sin \alpha }}} - \sqrt {\frac{{1 + \sin \alpha }}{{1 - \sin \alpha }}} =$
- A$tan\,\,\alpha$
- ✓$2\,\, tan\,\,\alpha$
- C$2\,\, cot\,\,\alpha$
- D$cot\,\,\alpha$
${=\frac{(1-\sin \alpha)-(1+\sin \alpha)}{\sqrt{1-\sin ^{2} \alpha}}} $
${=\frac{-2 \sin \alpha}{|\cos \alpha|}=\frac{-2 \sin \alpha}{-\cos \alpha}} $
$ \left[\because {\frac{\pi }{2} < \alpha < \pi \therefore \cos \alpha {\rm{ is - ve }}} \right] = 2{\rm{tan}}\alpha $
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Statement $-1$:An equation of a common tangent to these curve is $y = x + \sqrt 5 $
Statement $-2$: If the line, $y = mx + \frac{{\sqrt 5 }}{m}\left( {m \ne 0} \right)$ is their common tangent , then $m$ satisfies ${m^4} - 3{m^2} + 2 = 0$.