A) \[\frac{x}{\sqrt{1-{{x}^{2}}}}\]
B) \[\frac{1-2x}{\sqrt{1-{{x}^{2}}}}\]
C) \[\frac{1-2x}{2\sqrt{1-{{x}^{2}}}}\]
D) \[\frac{1}{1+{{x}^{2}}}\]
Correct Answer: C
Solution :
Put \[x=\cos \theta \] \[\therefore y={{\tan }^{-1}}\frac{\cos \theta }{1+\sin \theta }+\sin \left[ 2{{\tan }^{-1}}\sqrt{\left( \frac{1-\cos \theta }{1+\cos \theta } \right)} \right]\] \[={{\tan }^{-1}}\frac{\sin \varphi }{1+\cos \varphi }+\sin \left[ 2{{\tan }^{-1}}\tan \left( \frac{\theta }{2} \right) \right]\], {where \[\varphi =90{}^\circ -\theta \]} \[={{\tan }^{-1}}\tan \left( \frac{\varphi }{2} \right)+\sin \left( 2.\frac{\theta }{2} \right)=\left( \frac{\varphi }{2} \right)+\sin \theta \] \[=\frac{\pi }{4}-\frac{\theta }{2}+\sqrt{1-{{\cos }^{2}}\theta }=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x+\sqrt{1-{{x}^{2}}}\] \[\therefore \frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{1-{{x}^{2}}}}+\frac{1}{2\sqrt{1-{{x}^{2}}}}(-2x)=\frac{1-2x}{2\sqrt{1-{{x}^{2}}}}\]
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