Question

Prove that
$2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$

Answer 1

Hint: For solving this question we will use the formula $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ , ${{\tan }^{2}}\theta \text{=se}{{\text{c}}^{2}}\theta \text{-1}$ and ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ . After that, we will try to solve $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ and prove that, it is equal to ${{\tan }^{-1}}\left( 1 \right)$ . Then, we will prove the desired result easily.

Complete step-by-step solution -
Given:
We have to prove the following equation:
$2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$
Now, let $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ , $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ and $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . So, we will solve for $\alpha$ , $\beta$ and $\gamma$ separately. Then, we will find the value of $\alpha +\beta +\gamma$ .
Now, before we proceed further we should know the following formulas:
$2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)\text{ }\left( \text{if }-1 < x < 1 \right)...............\left( 1 \right)$
${{\tan }^{2}}\theta \text{=se}{{\text{c}}^{2}}\theta \text{-1 }....................\left( 2 \right)$
${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\text{ }\left( \text{if }xy<1 \right).............................\left( 3 \right)$
${{\tan }^{-1}}1=\dfrac{\pi }{4}..................................\left( 4 \right)$
Now, we will simplify $\alpha$ , $\beta$ and $\gamma$ separately with the help of the above four formulas.
Simplification of $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ :
Now, we will use the formula from the equation (1) as $-1 < \dfrac{1}{5} < 1$ to write $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{5}{12} \right)$ . Then,
$\begin{aligned} & \alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right) \\\ & \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{5}}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right) \\\ & \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{2\times 5}{{{5}^{2}}-1} \right) \\\ & \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{10}{25-1} \right) \\\ & \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{10}{24} \right) \\\ & \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{5}{12} \right)........................\left( 5 \right) \\\ \end{aligned}$
Simplification of $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ :
Now, we will use the formula from the equation (1) as $-1<\dfrac{1}{8}<1$ to write $2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{16}{63} \right)$ . Then,
$\begin{aligned} & \gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\\ & \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{8}}{1-{{\left( \dfrac{1}{8} \right)}^{2}}} \right) \\\ & \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{2\times 8}{{{8}^{2}}-1} \right) \\\ & \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{16}{64-1} \right) \\\ & \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{16}{63} \right)........................\left( 6 \right) \\\ \end{aligned}$
Simplification of $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ :
Now, we write $\sec \beta =\dfrac{5\sqrt{2}}{7}$ and apply the formula from the equation (2) to get the value of $\tan \beta$ . Then,
$\begin{aligned} & {{\tan }^{2}}\beta ={{\sec }^{2}}\beta -1 \\\ & \Rightarrow {{\tan }^{2}}\beta ={{\left( \dfrac{5\sqrt{2}}{7} \right)}^{2}}-1 \\\ & \Rightarrow {{\tan }^{2}}\beta =\dfrac{50}{49}-1 \\\ & \Rightarrow {{\tan }^{2}}\beta =\dfrac{1}{49} \\\ & \Rightarrow \tan \beta =\sqrt{\dfrac{1}{49}} \\\ & \Rightarrow \tan \beta =\dfrac{1}{7} \\\ \end{aligned}$
Now, as $\dfrac{5\sqrt{2}}{7}>1$ so, we can say that, $0<{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)<\dfrac{\pi }{2}$ and as $\tan \beta =\dfrac{1}{7}$ so, we can also write $\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)$ . Then,
$\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)..........................\left( 7 \right)$
Now, we will calculate the value of $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . And as per our assumption $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ , $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ and $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . Then,
$\begin{aligned} & 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\\ & \Rightarrow \alpha +\beta +\gamma \\\ \end{aligned}$
Now, we put $\alpha ={{\tan }^{-1}}\left( \dfrac{5}{12} \right)$ from equation (5), $\gamma ={{\tan }^{-1}}\left( \dfrac{16}{63} \right)$ from equation (6) and $\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)$ from equation (7) in the above expression. Then,
$\begin{aligned} & \alpha +\beta +\gamma \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\\ \end{aligned}$
Now, we will use the formula from equation (3) as $\dfrac{5}{12}\times \dfrac{1}{7}=\dfrac{5}{84}<1$ to write ${{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)={{\tan }^{-1}}\left( \dfrac{47}{79} \right)$ in the above line. Then,
$\begin{aligned} & {{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{5}{12}+\dfrac{1}{7}}{1-\dfrac{5}{12}\times \dfrac{1}{7}} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{5\times 7+12}{7\times 12-5} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{35+12}{84-5} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\\ \end{aligned}$
Now, we will use the formula from equation (3) as $\dfrac{47}{79}\times \dfrac{16}{63}=\dfrac{752}{4788}<1$ to write ${{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right)={{\tan }^{-1}}\left( 1 \right)$ in the above line. Then,
$\begin{aligned} & {{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{47}{79}+\dfrac{16}{63}}{1-\dfrac{47}{79}\times \dfrac{16}{63}} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{47\times 63+16\times 79}{79\times 63-47\times 16} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{2961+1264}{4977-752} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( \dfrac{4225}{4225} \right) \\\ & \Rightarrow {{\tan }^{-1}}\left( 1 \right) \\\ \end{aligned}$
Now, from the formula from the equation (4), we can write ${{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ . Then,
$\begin{aligned} & {{\tan }^{-1}}\left( 1 \right) \\\ & \Rightarrow \dfrac{\pi }{4} \\\ \end{aligned}$
Now, from the above result, we conclude that, $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$ .
Thus, $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$ .
Hence, proved.

Note: Here, the student should understand what is asked in the question and then proceed in the right direction to prove the desired result quickly. Moreover, we should use formulas ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ and ${{\tan }^{2}}\theta \text{=se}{{\text{c}}^{2}}\theta \text{-1}$ to avoid tough calculation. And avoid calculation mistakes while solving.