A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{\sqrt{3}}{2}\]
C) \[\frac{1}{2}\]
D) \[\frac{\pi }{4}\]
Correct Answer: A
Solution :
\[\cos \,\left[ {{\tan }^{-1}}\frac{1}{3}+{{\tan }^{-1}}\frac{1}{2} \right]=\cos \,\left[ {{\tan }^{-1}}\left( \frac{\frac{1}{3}+\frac{1}{2}}{1-\frac{1}{3}\times \frac{1}{2}} \right) \right]\] \[=\cos \,\{{{\tan }^{-1}}(1)\}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\].
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