Inverse Trigonometric Functions Question 105
Question: The value of $ {{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18 $ is
Options:
A) $ \pi $
B) $ \frac{\pi }{2} $
C) $ {{\cot }^{-1}}5 $
D) $ {{\cot }^{-1}}3 $
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Answer:
Correct Answer: D
We have $ {{\cot }^{-1}}7+{{\cot }^{-1}}8+{{\cot }^{-1}}18 $
$ {{\tan }^{-1}}\frac{1}{7}+{{\tan }^{-1}}\frac{1}{8}+{{\tan }^{-1}}\frac{1}{18} $
$ ={{\tan }^{-1}}( \frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7}\times \frac{1}{8}} )+{{\tan }^{-1}}\frac{1}{18} $
$ ={{\tan }^{-1}}\frac{15}{55}{{\tan }^{-1}}\frac{1}{18}( \because \frac{1}{7}.\frac{1}{8}<1 ) $
$ {{\tan }^{-1}}\frac{3}{11}+{{\tan }^{-1}}\frac{1}{18}={{\tan }^{-1}}( \frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11}\times \frac{1}{18}} ) $
$ ( \because \frac{3}{11}.\frac{1}{18}<1 ) $
$ ={{\tan }^{-1}}\frac{65}{195}={{\tan }^{-1}}\frac{1}{3}={{\cot }^{-1}}3 $
BETA
