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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
$\cot ^{-1}\left(\frac{a b+1}{a-b}\right)+\cot ^{-1}\left(\frac{b c+1}{b-c}\right)+\cot ^{-1}\left(\frac{c a+1}{c-a}\right)$ is equal to
  • ✓
    $0$
  • B
    $1$
  • C
    $\frac{\pi}{4}$
  • D
    none of these
✓

Answer

Correct option: A.
$0$
(A) Since, $\cot ^{-1} x-\cot ^{-1} y=\cot ^{-1}\left(\frac{x y+1}{y-x}\right)$
$\therefore \quad \cot ^{-1} \frac{a b+1}{a-b}+\cot ^{-1} \frac{b c+1}{b-c}+\cot ^{-1} \frac{c a+1}{c-a}$
$=\cot ^{-1} b-\cot ^{-1} a+\cot ^{-1} c-\cot ^{-1} b$ $+\cot ^{-1} a -\cot ^{-1} c$
$=0$

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