A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{3}\]
D) None of these
Correct Answer: B
Solution :
Suppose, \[A=\frac{\pi }{6}\] and \[B=\frac{\pi }{4}\] Now, \[\sin A={{\sin }^{2}}B\] \[\Rightarrow \] \[sin\frac{\pi }{6}={{\sin }^{2}}\left( \frac{\pi }{4} \right)\] \[\Rightarrow \] \[\frac{1}{2}={{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\] \[\Rightarrow \] \[\frac{1}{2}=\frac{1}{2}\] (true) Now, \[2{{\cos }^{2}}A=3{{\cos }^{2}}B\] \[\therefore \] \[2{{\cos }^{2}}\frac{\pi }{6}=3{{\cos }^{2}}\left( \frac{\pi }{4} \right)\] \[\Rightarrow \] \[2\times {{\left( \frac{\sqrt{3}}{2} \right)}^{2}}=3{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}\] \[\Rightarrow \] \[2\times \frac{3}{4}=3\times \frac{1}{2}\] \[\Rightarrow \] \[\frac{3}{2}=\frac{3}{2}\] (true)
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