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prove that $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\cot ^{-1} 2$

written 5.2 years ago by teamques10 ★ 68k

modified 2.6 years ago by pedsangini276 • 4.8k

engineering mathematics

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written 5.2 years ago by teamques10 ★ 68k

Solution:

$\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$

$\quad=\tan ^{-1}\left[\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right]$

\begin{aligned} &=\tan ^{-1}\left(\frac{1}{2}\right) \ &=\cot ^{-1} 2 \ \therefore & \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\cot ^{-1} 2 \end{aligned}

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