A) \[{{\tan }^{-1}}\left( \frac{\pi }{2}-1 \right)\]
B) \[{{\tan }^{-1}}\left( \frac{\pi }{2}+1 \right)\]
C) \[2{{\tan }^{-1}}\left( \frac{\pi }{2}-1 \right)\]
D) \[-2{{\tan }^{-1}}\left( \frac{\pi }{2}-1 \right)\]
Correct Answer: A
Solution :
[a]: \[x+y=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx},\frac{dz}{dx}=1+\cos z,\] \[dx=\frac{dz}{1+\cos z}=\frac{1}{2}{{\sec }^{2}}\frac{z}{2}dz\] \[\Rightarrow \]\[x=\tan \frac{z}{2}+c\]or\[x=\tan \left( \frac{x+y}{2} \right)+c\] \[y\left( \frac{\pi }{2} \right)=0\Rightarrow \frac{\pi }{2}=1+c\Rightarrow c=\frac{\pi }{2}-1\] \[x=\tan \left( \frac{x+y}{2} \right)+\frac{\pi }{2}-1\] \[y=-x+2{{\tan }^{-1}}\left( x+1-\frac{\pi }{2} \right)\] \[y(0)=2{{\tan }^{-1}}\left( 1-\frac{\pi }{2} \right)=-2{{\tan }^{-1}}\left( \frac{\pi }{2}-1 \right)\].
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