JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(\alpha\gt\beta\gt\gamma\gt0\), then the expression \(\cot ^{-1}\left\{\beta+\frac{\left(1+\beta^2\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^2\right)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^2\right)}{(\gamma-\alpha)}\right\}\) is equal to :
- A \(\pi\)
- B 0
- C \(\frac{\pi}{2}-(\alpha+\beta+\gamma)\)
- D \(3 \pi\)
Answer & Solution
Correct Answer
(A) \(\pi\)
Step-by-step Solution
Detailed explanation
\(\Rightarrow \cot ^{-1}\left(\frac{\alpha \beta+1}{\alpha-\beta}\right)+\cot ^{-1}\left(\frac{\beta \gamma+1}{\beta-\gamma}\right)+\cot ^{-1}\left(\frac{\alpha \gamma+1}{\gamma-\alpha}\right) \)…
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