A) 0
B) \[\frac{\pi ab+c(b-c)}{a+b}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi ab+c(a-b)}{a+b}\]
Correct Answer: D
Solution :
[d] \[a{{\sin }^{-1}}x-b{{\cos }^{-1}}x=c\] We have b \[{{\sin }^{-1}}x+b{{\cos }^{-1}}x=\frac{b\pi }{2}\] Adding \[(a+b)\]\[{{\sin }^{-1}}x=\frac{b\pi }{2}+c\] Or \[{{\sin }^{-1}}x=\frac{\left( \frac{b\pi }{2} \right)+c}{a+b}=\frac{b\pi +2c}{2(a+b)}\] \[\therefore {{\cos }^{-1}}x=\frac{\pi }{2}-\frac{b\pi +2c}{2(a+b)}=\frac{\pi a-2c}{2(a+b)}\] \[\Rightarrow a{{\sin }^{-1}}x+b{{\cos }^{1}}x=\frac{\pi ab+c(a-b)}{a+b}\]
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