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Qn #1814
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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:
A.
$\dfrac{n}{n+1}$
B.
$\dfrac{n+1}{n+2}$
C.
$\dfrac{n}{n+2}$
D.
$\dfrac{n-1}{n+2}$
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❓ Question: If $ \theta = \tan^{-1}\dfrac{1}{1+2} + \tan^{-1}\... | TANCET Group Studies | Tancet Group Studies