MCQ
$\int_0^{\pi /2} {\frac{{\cos x - \sin x}}{{1 + \sin x\cos x}}} \,dx = $
- A$2$
- B$ - 2$
- ✓$0$
- DNone of these
Now $I = \int_0^{\pi /2} {\frac{{\cos \left( {\frac{\pi }{2} - x} \right) - \sin \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}\,dx} $
$= \int_0^{\pi /2} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}\,\,dx} $.....$(ii)$
On adding, $2I = 0 \Rightarrow I = 0$.
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$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5} \text { and } \frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$ is :