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Question

Suppose tanα=14\tan \alpha=\frac{1}{4} and tanβ=35\tan \beta=\frac{3}{5}, where 0<α<β<π20<\alpha<\beta<\frac{\pi}{2}. Find tan(α+β)\tan (\alpha+\beta). Then show that Tan114+Tan135=π4\operatorname{Tan}^{-1} \frac{1}{4}+\operatorname{Tan}^{-1} \frac{3}{5}=\frac{\pi}{4}.

Solution

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We will use the formula: tan(α+β)=tanα+tanβ1tanαtanβ\tan (\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta }{1- \tan \alpha\tan \beta}:

tan(α+β)=tanα+tanβ1tanαtanβ=14+3511435=17201320=17201720=1\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(α+β)=1\tan (\alpha + \beta) = 1, so α+β=Tan11=π4\alpha + \beta = \text{Tan}^{-1} 1 = \dfrac{\pi}{4}.

Now, notice that if tanα=14\tan \alpha = \dfrac{1}{4}, then α=Tan114\alpha = \text{Tan}^{-1} \dfrac{1}{4}. Similarly, if tanβ=35\tan \beta = \dfrac{3}{5}, then β=Tan135\beta = \text{Tan}^{-1} \dfrac{3}{5}. Therefore, Tan114+Tan135=α+β=π4\text{Tan}^{-1} \dfrac{1}{4}+ \text{Tan}^{-1} \dfrac{3}{5} = \alpha + \beta = \dfrac{\pi}{4}.

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