Question

Write the value of ${\tan }^{-1}\frac{a}{b}-{\tan }^{-1}\left(\frac{a-b}{a+b}\right)$.

Answer 1

We know that ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$.
Now,
$$\begin{gathered} {\tan }^{-1}\frac{a}{b}-{\tan }^{-1}\left(\frac{a-b}{a+b}\right)={\tan }^{-1}\left(\frac{\frac{a}{b}-\frac{a-b}{a+b}}{1+\frac{a}{b}\frac{a-b}{a+b}}\right) \\ ={\tan }^{-1}\left(\frac{\frac{a^2+ab-ab+b^2}{b\left(a+b\right)}}{\frac{ab+b^2-ab+a^2}{b\left(a+b\right)}}\right) \\ ={\tan }^{-1}\left(1\right) \\ ={\tan }^{-1}\left(\tan \frac{\mathrm{\pi }}{4}\right)\left[\because \tan \frac{\mathrm{\pi }}{4}=1\right] \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

∴ ${\tan }^{-1}\frac{a}{b}-{\tan }^{-1}\left(\frac{a-b}{a+b}\right)=\frac{\mathrm{\pi }}{4}$