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Question

# Suppose $\tan \alpha=\frac{1}{4}$ and $\tan \beta=\frac{3}{5}$, where $0<\alpha<\beta<\frac{\pi}{2}$. Find $\tan (\alpha+\beta)$. Then show that $\operatorname{Tan}^{-1} \frac{1}{4}+\operatorname{Tan}^{-1} \frac{3}{5}=\frac{\pi}{4}$.

Solution

Verified

We will use the formula: $\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta }{1- \tan \alpha\tan \beta}$:

\begin{align*} \tan (\alpha + \beta) &= \dfrac{\tan \alpha + \tan \beta }{1- \tan \alpha\tan \beta}\\ &=\dfrac{\frac{1}{4}+\frac{3}{5}}{1-\frac{1}{4}\cdot\frac{3}{5}}\\ &=\dfrac{\frac{17}{20}}{1-\frac{3}{20}}\\ &=\dfrac{\frac{17}{20}}{\frac{17}{20}}\\ &=1 \end{align*}

Since $\tan (\alpha + \beta) = 1$, so $\alpha + \beta = \text{Tan}^{-1} 1 = \dfrac{\pi}{4}$.

Now, notice that if $\tan \alpha = \dfrac{1}{4}$, then $\alpha = \text{Tan}^{-1} \dfrac{1}{4}$. Similarly, if $\tan \beta = \dfrac{3}{5}$, then $\beta = \text{Tan}^{-1} \dfrac{3}{5}$. Therefore, $\text{Tan}^{-1} \dfrac{1}{4}+ \text{Tan}^{-1} \dfrac{3}{5} = \alpha + \beta = \dfrac{\pi}{4}$.

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