Question

Prove the result $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{4}{7} \right)$.

Answer 1

Hint:We will apply the formula $2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ in order to solve the question. We will apply the trick of first considering the left hand side and then verify it to the right hand side of the expression. We will also use ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ to solve the question further.

Complete step-by-step answer:
We will start solving the question by taking the expression $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{4}{7} \right)...(i)$ into consideration. We will focus on the left hand side of the expression which is given by $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$. Now we will apply the formula of trigonometry. The formula is as $2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$. Therefore the left hand side of the expression changes into
$\begin{aligned} & 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{2\left( \dfrac{1}{5} \right)}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\\ & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{2}{1}}{\dfrac{24}{5}} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\\ & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{2}{1}\times \dfrac{5}{24} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\\ & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\\ \end{aligned}$
Now we will apply the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$. Therefore, we have
$\begin{aligned} & 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\\ & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{5}{12}+\dfrac{1}{8}}{1-\left( \dfrac{5}{12} \right)\left( \dfrac{1}{8} \right)} \right) \\\ & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{10+3}{24}}{\dfrac{96-5}{96}} \right) \\\ & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{13}{24}}{\dfrac{91}{96}} \right) \\\ & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{13}{24}\times \dfrac{96}{91} \right) \\\ & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{4}{7} \right) \\\ \end{aligned}$
And since it is equal to the right hand side of the expression (i). Therefore the two given sides are equal.
Hence, we have proved $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{4}{7} \right)$.

Note: We can also split $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ into ${{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ instead of using the formula $2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and applying the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ to it. This may ease in solving the question. Be aware of doing calculation mistakes otherwise the answer will be wrong and the verification will fail. While solving ${{\tan }^{-1}}\left( \dfrac{\dfrac{2}{1}}{\dfrac{24}{5}} \right)$ write ${{\tan }^{-1}}\left( \dfrac{2}{1}\times \dfrac{5}{24} \right)$ instead of ${{\tan }^{-1}}\left( \dfrac{2}{1}\times \dfrac{24}{5} \right)$. This is due to the fact that $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$. That is why we have done calculation in this manner.