A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{6}\]
D) None of these
Correct Answer: A
Solution :
\[{{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{3}{5}-{{\tan }^{-1}}\frac{8}{19}\] \[={{\tan }^{-1}}\left[ \frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4}\times \frac{3}{5}} \right]-{{\tan }^{-1}}\frac{8}{19}={{\tan }^{-1}}\frac{27}{11}-{{\tan }^{-1}}\frac{8}{19}\] \[={{\tan }^{-1}}\left[ \frac{\frac{27}{11}-\frac{8}{19}}{1+\frac{27}{11}\times \frac{8}{19}} \right]={{\tan }^{-1}}\left( \frac{425}{425} \right)={{\tan }^{-1}}(1)=\frac{\pi }{4}\].
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