A) 0
B) 1/2
C) \[-1/2\]
D) 1
Correct Answer: B
Solution :
Given, \[4{{\sin }^{-1}}x+{{\cos }^{-1}}x=\pi \] \[\Rightarrow \] \[4{{\sin }^{-1}}x+\frac{\pi }{2}-{{\sin }^{-1}}x=\pi \] \[\left( \because \,{{\cos }^{-1}}x+{{\sin }^{-1}}x=\frac{\pi }{2} \right)\] \[\Rightarrow \] \[3{{\sin }^{-1}}x=\pi -\frac{\pi }{2}\] \[\Rightarrow \] \[3{{\sin }^{-1}}x=\frac{\pi }{2}\] \[\Rightarrow \] \[x=\sin \frac{\pi }{6}\] \[\Rightarrow \] \[x=\frac{1}{2}\]
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