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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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