A) \[\alpha \]
B) \[2\alpha \]
C) \[-\alpha \]
D) \[-2\alpha \]
Correct Answer: A
Solution :
Roots of equation \[{{\operatorname{x}}^{2}}- 9x+ 8 = 0\] are 1 and 8. Let \[\operatorname{y} = \left[ si{{n}^{2}}\alpha + si{{n}^{4}}\alpha + si{{n}^{6}}\alpha +...\,\infty \right] {{\log }_{e}}2\] \[\Rightarrow \,\,y=\frac{{{\sin }^{2}}\alpha }{1-{{\sin }^{2}}\alpha }\,{{\log }_{e}}2=\,\,ta{{n}^{2}}\alpha \,{{\log }_{e}}2\] \[\Rightarrow \,\,\,y={{\log }_{e}}2{{\,}^{{{\tan }^{2}}\alpha }}\] \[\Rightarrow \,\,\,{{e}^{y}}={{2}^{{{\tan }^{2}}\alpha }}\] According to question, \[{{2}^{{{\tan }^{2}}\alpha }}\,\,=\,\,8={{2}^{3}}\Rightarrow {{\tan }^{2}}\alpha \,\,=\,\,3\] \[\Rightarrow \,\,\,tan\,\alpha =\sqrt{3}\,\,\Rightarrow \,\,\alpha =\frac{\pi }{3}\] \[\therefore \,\,{{\sin }^{-1}}\left( \sin \frac{2\pi }{3} \right)=\pi -\frac{2\pi }{3}=\frac{\pi }{3}=\alpha \]
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