Aspire Faculty ID #16410 · Topic: NIMCET 2009 · Just now
NIMCET 2009

If $ \theta = \tan^{-1}\dfrac{1}{1+2} + \tan^{-1}\dfrac{1}{1+2\cdot3} + \tan^{-1}\dfrac{1}{1+3\cdot4} + \ldots + \tan^{-1}\dfrac{1}{1+n(n+1)} $, then $\tan\theta$ is equal to:

Solution

We use identity $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\dfrac{a-b}{1+ab}$ 
Here each term satisfies 
$\tan^{-1}\dfrac{1}{1+k(k+1)} = \tan^{-1}(k+1) - \tan^{-1}(k)$ 
 So the series telescopes: 
$\theta = \tan^{-1}(n+1) - \tan^{-1}(1)$ 
 Thus 
$\tan\theta = \dfrac{(n+1)-1}{1+(n+1)(1)} = \dfrac{n}{n+2}$

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