Inverse Trigonometric Functions Question 54
Question: The sum to the n term of the series $ \cos e{c^{-1}}\sqrt{10}+\cos e{c^{-1}}\sqrt{50}+\cos e{c^{-1}}\sqrt{170}+… $
$ +\cos e{c^{-1}}\sqrt{(n^{2}+1)(n^{2}+2n+2)} $
Options:
A) $ {{\tan }^{-1}}(n+1)-\pi /4 $
B) $ \pi /4 $
C) $ {{\tan }^{-1}}(n+1) $
D) 1
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Answer:
Correct Answer: A
Let $ \theta =\cos e{c^{-1}}\sqrt{(n^{2}+1)(n^{2}+2n+2)} $
$ \Rightarrow \cos ec^{2}\theta =(n^{2}+1)(n^{2}+2n+2) $
$ ={{(n^{2}+1)}^{2}}+2n(n^{2}+1)+n^{2}+1 $
$ ={{(n^{2}+n+1)}^{2}}+1\Rightarrow {{\cot }^{2}}\theta ={{(n^{2}+n+1)}^{2}} $
$ \Rightarrow \tan \theta =\frac{1}{n^{2}+n+1}=\frac{(n+1)-n}{1+(n+1)n} $
$ \Rightarrow \theta ={{\tan }^{-1}}[ \frac{(n+1)-n}{1+(n+1)n} ]={{\tan }^{-1}}(n+1)-ta{n^{-1}}n $
Thus, sum n terms of the given series $ =(ta{n^{-1}}2-ta{n^{-1}}1)+(ta{n^{-1}}3-ta{n^{-1}}2) $
$ +(ta{n^{-1}}4-ta{n^{-1}}3)+…+(ta{n^{-1}}(n+1)-ta{n^{-1}}n) $
$ \Rightarrow {{\tan }^{-1}}(n+1)-\pi /4 $
BETA
