A) \[\sin 2\beta \]
B) \[\sin 4\beta \]
C) \[\sin 3\beta \]
D) None of these
Correct Answer: B
Solution :
\[\cos 2\alpha =\frac{1-{{t}^{2}}}{1+{{t}^{2}}}=\frac{24}{25}\]{Here\[t=\tan \alpha \]} \[\sin 2\beta =\frac{2T}{1+{{T}^{2}}}=\frac{3}{5}\Rightarrow \cos 2\beta =\frac{4}{5}\]{\[T=\tan \beta \]} \[\therefore \,\,\sin 4\beta =2\sin 2\beta \cos 2\beta =2.\frac{3}{5}.\frac{4}{5}=\frac{24}{25}=\cos 2\alpha \].
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