Trigonometry & Inverse Trigonometry
Telescoping sum via cot⁻¹ to tan⁻¹ conversion
nta_pyq_2025_apr
Grade 12
Question:
If $\alpha>\beta>\gamma>0$, then the expression $\cot^{-1}\!\left\{\beta+\dfrac{1+\beta^2}{\alpha-\beta}\right\}+\cot^{-1}\!\left\{\gamma+\dfrac{1+\gamma^2}{\beta-\gamma}\right\}+\cot^{-1}\!\left\{\alpha+\dfrac{1+\alpha^2}{\gamma-\alpha}\right\}$ is equal to:
$\pi$
$0$
$\dfrac{\pi}{2}-(\alpha+\beta+\gamma)$
$3\pi$
Step-by-Step Solution
Key Concept: Simplify each term to $\cot^{-1}\!\left(\dfrac{1+xy}{x-y}\right)=\tan^{-1}x-\tan^{-1}y$ (for $x>y>0$), creating a telescoping sum; the third term requires adding $\pi$ because the denominator is negative.
Each argument simplifies: first term $\Rightarrow \cot^{-1}\!\dfrac{\alpha\beta+1}{\alpha-\beta}=\tan^{-1}\alpha-\tan^{-1}\beta$.
Second term $\Rightarrow \tan^{-1}\beta-\tan^{-1}\gamma$.
Third term (denominator $\gamma-\alpha<0$): $\cot^{-1}\!\dfrac{\alpha\gamma+1}{\gamma-\alpha}=\pi+\tan^{-1}\gamma-\tan^{-1}\alpha$.
Sum $=(\tan^{-1}\alpha-\tan^{-1}\beta)+(\tan^{-1}\beta-\tan^{-1}\gamma)+(\pi+\tan^{-1}\gamma-\tan^{-1}\alpha)=\pi$.
Correct Answer: 1