Question

Prove the following expression $2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)=\dfrac{\pi }{4}$.

Answer 1

Hint:We will use the formulas of trigonometry to which we are familiar. The ones that are used here are $\sin x=\dfrac{\text{perpendicular}}{\text{Hypotenuse}}$ and $\tan x=\dfrac{\text{perpendicular}}{\text{Base}}$ for finding the values of trigonometric terms. Also, we have used ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ to solve the question further.

Complete step-by-step answer:
Let us first consider the expression $2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)=\dfrac{\pi }{4}...(i)$. We will start with substituting ${{\sin }^{-1}}\left( \dfrac{3}{5} \right)=x$. Thus after placing the inverse sine term to the right side of the equation we will get $\dfrac{3}{5}=\sin x$. By the formula $\sin x=\dfrac{\text{perpendicular}}{\text{Hypotenuse}}$ we have 3 as perpendicular and 5 as hypotenuse. The diagram for this is given as

Now we will apply the Pythagoras theorem. Thus we will get,
$\begin{align}
  & {{\left( 5 \right)}^{2}}={{y}^{2}}+{{\left( 3 \right)}^{2}} \\
 & \Rightarrow 25={{y}^{2}}+9 \\
 & \Rightarrow {{y}^{2}}=25-9 \\
 & \Rightarrow {{y}^{2}}=16 \\
 & \Rightarrow {{y}^{2}}=\sqrt{16} \\
 & \Rightarrow y=\pm 4 \\
\end{align}$
As the side of the triangle cannot be negative therefore y = 4. Now we will use the formula given by $\tan x=\dfrac{\text{perpendicular}}{\text{Base}}$. As we have perpendicular as 3 and base as 4. Thus we have $\tan \left( x \right)=\dfrac{3}{4}$. Taking the inverse tangent to the right side of the equation we will have $x={{\tan }^{-1}}\left( \dfrac{3}{4} \right)$. Since, we have ${{\sin }^{-1}}\left( \dfrac{3}{5} \right)=x$ thus, ${{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\tan }^{-1}}\left( \dfrac{3}{4} \right)$. Now, we will substitute this value in the left hand side of the equation (i) which results into $\begin{align}
  & 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)=2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right) \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right) \\
\end{align}$
Now, we will apply the formula given by ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$. Thus our equation changes into
$\begin{align}
  & 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{3}{4} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right) \\
 & \Rightarrow \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{4}+\dfrac{3}{4}}{1-\left( \dfrac{3}{4} \right)\left( \dfrac{3}{4} \right)} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right) \\
 & \Rightarrow \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{6}{4}}{\dfrac{16-9}{16}} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right) \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{6}{1}\times \dfrac{4}{7} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right) \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{27}{7} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right) \\
\end{align}$
Now we will again apply the formula given by ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$. Thus, we get
$\begin{align}
  & 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{24}{7}-\dfrac{17}{31}}{1+\left( \dfrac{24}{7} \right)\left( \dfrac{17}{31} \right)} \right) \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{744-119}{217}}{\dfrac{217+408}{217}} \right) \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \dfrac{645}{645} \right) \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( 1 \right) \\
\end{align}$
As we know that the value of ${{\tan }^{-1}}\left( \dfrac{\pi }{4} \right)=1$ and ${{\tan }^{-1}}\tan \left( x \right)=x$ we can have
$\begin{align}
  & 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( 1 \right) \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4} \right) \right) \\
 & \Rightarrow 2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)=\dfrac{\pi }{4} \\
\end{align}$
Hence, the value of the expression $2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{17}{31} \right)=\dfrac{\pi }{4}$.

Note: Alternatively we could have used the formula given by $\tan \left( 2x \right)=\dfrac{2\tan \left( x \right)}{1-{{\tan }^{2}}\left( x \right)}$ after $\tan \left( x \right)=\dfrac{3}{4}$. By this we can directly do the substitution $2{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$ in expression (i). Also, instead of splitting $2{{\tan }^{-1}}\left( \dfrac{3}{4} \right)$ into ${{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{3}{4} \right)$ we could have used the formula of $2{{\tan }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ and proceeded further. Moreover, instead of converting the inverse sine term into tangent with the help of diagram and Pythagoras we could have use direct formula for converting it. The formula for this is given by ${{\sin }^{-1}}x={{\tan }^{-1}}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)$.