A) Is \[{{\tan }^{-1}}\left( \frac{1}{9} \right)\]
B) Is \[{{\tan }^{-1}}\frac{1}{40}\]
C) Cannot be evaluated
D) Is\[2{{\tan }^{-1}}\left( 1/9 \right)\]
E) Is \[2{{\tan }^{-1}}\frac{1}{40}\]
Correct Answer: D
Solution :
\[\because \] \[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\]; \[\frac{5}{\sin \left( \frac{\pi }{2}+B \right)}=\frac{4}{\sin B}\] \[\frac{5}{\cos B}=\frac{4}{\sin B}\], \ \[\tan B=\frac{4}{5}\] \[\tan A=\tan \left( \frac{\pi }{2}+B \right)=-\cot B=\frac{-5}{4}\] \[\tan C=\tan (\pi -(A+B))\]\[=-\tan (A+B),\]\[[A+B+C=\pi ]\] \[=-\frac{(\tan A+\tan B)}{1-\tan A.\tan B}\]\[=\frac{-\left( -\frac{5}{4}+\frac{4}{5} \right)}{1+1}=\frac{9}{40}\] \[C={{\tan }^{-1}}\left( \frac{\left( 2.\frac{1}{9} \right)}{1-{{\left( \frac{1}{9} \right)}^{2}}} \right)\]; \ \[C=2{{\tan }^{-1}}\left( \frac{1}{9} \right)\].
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