Aspire Faculty ID #3628 · Topic: NIMCET 2012 · Just now
NIMCET 2012

The value of $\cot^{-1}(21) + \cot^{-1}(13) + \cot^{-1}(-8)$ is:

Solution

Use identity for inverse cotangent: 
$\cot^{-1}a + \cot^{-1}b = \cot^{-1}\left(\dfrac{ab - 1}{a + b}\right)$ 
Apply to the first two terms: 
$\cot^{-1}(21) + \cot^{-1}(13)$ 
$= \cot^{-1}\left(\dfrac{21 \cdot 13 - 1}{21 + 13}\right)$ 
$= \cot^{-1}\left(\dfrac{273 - 1}{34}\right)$ 
$= \cot^{-1}\left(\dfrac{272}{34}\right)$ 
$= \cot^{-1}(8)$ 
Now the expression becomes: 
$\cot^{-1}(8) + \cot^{-1}(-8)$ 
But we know: $\cot^{-1}(x) + \cot^{-1}(-x) = \pi$ 
So, $\cot^{-1}(8) + \cot^{-1}(-8) = \pi$
$\therefore$ The value is $\pi$.

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