A) \[{{\tan }^{-1}}\frac{27}{11}\]
B) \[{{\sin }^{-1}}\frac{11}{27}\]
C) \[{{\cos }^{-1}}\frac{11}{27}\]
D) None of these
Correct Answer: A
Solution :
\[{{\cos }^{-1}}\frac{4}{5}+{{\tan }^{-1}}\frac{3}{5}={{\tan }^{-1}}\left[ \frac{\sqrt{\left( 1-\frac{16}{25} \right)}}{\frac{4}{5}} \right]+{{\tan }^{-1}}\frac{3}{5}\]\[\left[ \text{Since}\,\,{{\cos }^{-1}}x={{\tan }^{-1}}\frac{\sqrt{(1-{{x}^{2}})}}{x} \right]\] \[={{\tan }^{-1}}\frac{3}{4}+{{\tan }^{-1}}\frac{3}{5}={{\tan }^{-1}}\,\left( \frac{\frac{3}{4}+\frac{3}{5}}{1-\frac{3}{4}.\frac{3}{5}} \right)={{\tan }^{-1}}\left( \frac{27}{11} \right)\].
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