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Question

# ∀n∈N, x ∈R, the value of tan−1[x1.2+x2]+tan−1[x2.3+x2]+⋯+tan−1[xn(n+1)+x2], is

A
tan1[xn]tan1[xn+1]
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B
tan1[x]tan1[xn+1]
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C
tan1[n+1]tan1[x]
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D
tan1[x]
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Solution

## The correct option is B tan−1[x]−tan−1[xn+1]Let's assume rth term astr=tan−1(xr.(r+1)+x2)=tan−1⎛⎜ ⎜ ⎜⎝r+1x−rxrx.r+1x+1⎞⎟ ⎟ ⎟⎠We know that,tan−1(A−B1+AB)=tan−1A−tan−1BThus,tr=tan−1(r+1x)−tan−1(rx)Now add by Difference Method varying r from 1 to n, we getS=tan−1(n+1x)−tan−1(1x)Converting to cot−1 terms using tan−1x+cot−1x=π2, we getS=cot−1(1x)−cot−1(n+1x)Now, converting back to tan−1 terms by using tan−1(1x)=cot−1x, we getS=tan−1x−tan−1(xn+1)Hence, option B.

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