A) \[\frac{16}{15}\]
B) \[\frac{14}{15}\]
C) \[\frac{12}{15}\]
D) \[\frac{11}{15}\]
Correct Answer: B
Solution :
\[\sin \left[ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right]+\cos \,[{{\tan }^{-1}}(2\sqrt{2})]\] = \[\sin \left[ {{\tan }^{-1}}\frac{2/3}{1-1/9} \right]+\cos \,[{{\tan }^{-1}}(2\sqrt{2})]\] \[=\sin [{{\tan }^{-1}}3/4]+\cos \,[{{\tan }^{-1}}2\sqrt{2}]\] \[=\frac{3}{5}+\frac{1}{3}=\frac{14}{15}\].
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