Question

prove that:
|√1-sinQ÷1+sinQ+√1+sinQ÷1-sinQ|=-2÷cosQ , where π÷2 <Q<π​

Answer 1

$\sf \sqrt{ \frac{1 - \sin Q}{1 + \sin Q} } + \sqrt{ \frac{1 +\sin Q }{1 -\sin Q } }$

$\implies \sf \sqrt{ \frac{1 - \sin Q}{1 + \sin Q} \times \frac{1 - \sin Q}{1 - \sin Q} } + \sqrt{ \frac{1 +\sin Q }{1 -\sin Q } \times\frac{1 + \sin Q}{1 + \sin Q} }$

$\implies \sf \sqrt{ \frac{(1 - \sin Q )^{2} }{1 - \sin ^{2} Q} } + \sqrt{ \frac{(1 + \sin Q )^{2} }{1 -\sin ^{2} Q } }$

$\implies \sf \sqrt{ \frac{(1 - \sin Q )^{2} }{ \cos ^{2} Q} } + \sqrt{ \frac{(1 + \sin Q )^{2} }{\cos ^{2} Q } }$

$\implies \sf \sqrt{ (\frac{1 - \sin Q }{ \cos Q} ) ^{2} } + \sqrt{ (\frac{1 + \sin Q }{\cos Q }) ^{2} }$

$\implies \sf \frac{1 - \sin Q }{ \cos Q} + \frac{1 + \sin Q }{\cos Q }$

$\implies \sf \frac{1 - \sin Q + 1 + \sin Q }{ \cos Q}$

$\therefore \sf \frac{2 }{ \cos Q}$

$\sf since\: \frac{\pi}{2}<Q < \pi$

$\implies \sf \cos Q\: is\: negative$

$\orange{ \therefore \sf \frac{-2 }{ \cos Q} }$

HOPE IT HELPS!