Question

Let $\quad \alpha=\tan ^{-1}(1)+\frac{1}{2} \tan ^{-1}(2)+\frac{1}{3} \tan ^{-1}(3)$ and $\beta=\cot ^{-1}(1)+2 \cot ^{-1}(2)+3 \cot ^{-1}(3)$, then

• A. $6 \alpha-\cot ^{-1} 3=\frac{13 \pi}{4}$
• B. $\beta-\cot ^{-1} 3=\frac{3 \pi}{4}$
• C. $6 \alpha-\beta=\frac{5 \pi}{2}$
• D. $6 \alpha+\beta=4 \pi$

Answer 1

Answer: (A), (B), (C)

$6 \alpha=6 \cdot \frac{\pi}{4}+3 \tan ^{-1} 2+2 \tan ^{-1} 3$
$\therefore 6 \alpha-\cot ^{-1} 3=\frac{3 \pi}{2}+3 \cdot \frac{3 \pi}{4}-\frac{\pi}{2}=\pi+\frac{9 \pi}{4}=\frac{13 \pi}{4}$
and $\beta-\cot ^{-1} 3=\frac{\pi}{4}+\frac{2 \pi}{4}=\frac{3 \pi}{4}$.