NIMCET 2009 — Mathematics PYQ

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Question

NIMCET 2009 — Mathematics PYQ

NIMCET | Mathematics | 2009

If $θ = tan^{- 1} \frac{1}{1 + 2} + tan^{- 1} \frac{1}{1 + ( 2 ) ( 3 )} + tan^{- 1} \frac{1}{1 + ( 3 ) ( 4 )} + \dots + tan^{- 1} \frac{1}{1 + n ( n + 1 )}$ , then $tan θ$ is equal to:

Choose the correct answer:

Explanation

1. General Term Analysis

The $r^{th}$ term of the series, denoted as $T_r$ , can be written as:

T_r = \tan^{-1} \left( \frac{1}{1+r(r+1)} \right)

2. Applying the Identity

We know the identity: $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$ .

We can rewrite the numerator of the general term as $(r+1) - r$ :

T_r = \tan^{-1} \left( \frac{(r+1) - r}{1 + r(r+1)} \right)

T_r = \tan^{-1}(r+1) - \tan^{-1}(r)

3. Summing the Terms

The series for $\theta$ is the sum of terms from $r=1$ to $r=n$ :

\theta = \sum_{r=1}^{n} [\tan^{-1}(r+1) - \tan^{-1}(r)]

Expanding the summation:

For $r=1$ : $T_1 = \tan^{-1}(2) - \tan^{-1}(1)$
For $r=2$ : $T_2 = \tan^{-1}(3) - \tan^{-1}(2)$
For $r=3$ : $T_3 = \tan^{-1}(4) - \tan^{-1}(3)$
...
For $r=n$ : $T_n = \tan^{-1}(n+1) - \tan^{-1}(n)$

4. Telescoping Effect

When we add all terms, the intermediate terms cancel out:

\theta = [\tan^{-1}(2) - \tan^{-1}(1)] + [\tan^{-1}(3) - \tan^{-1}(2)] + \dots + [\tan^{-1}(n+1) - \tan^{-1}(n)]

\theta = \tan^{-1}(n+1) - \tan^{-1}(1)

5. Finding $\tan \theta$

Using the identity $\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left( \frac{x-y}{1+xy} \right)$ again:

\theta = \tan^{-1} \left( \frac{(n+1) - 1}{1 + (n+1)(1)} \right)

\theta = \tan^{-1} \left( \frac{n}{n+2} \right)

Therefore:

\tan \theta = \frac{n}{n+2}

Correct Option:

(c) $\frac{n}{n+2}$

Choose the correct answer:

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