Question

Prove that ${{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{3}{5}-{{\tan }^{-1}}\dfrac{8}{19}=\dfrac{\pi }{4}$ .

Answer 1

Hint: To solve this question first we consider the L.H.S. of the given expression which is ${{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{3}{5}-{{\tan }^{-1}}\dfrac{8}{19}$. Now, we find the value of L.H.S. by using the formula of inverse trigonometric functions ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy},\text{ if xy1}$. Now, we compare the obtained value of L.H.S. with R.H.S. to prove that both are equal.

Complete step-by-step answer:
We have to prove that ${{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{3}{5}-{{\tan }^{-1}}\dfrac{8}{19}=\dfrac{\pi }{4}$
We have been given an expression of inverse trigonometric function ${{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{3}{5}-{{\tan }^{-1}}\dfrac{8}{19}=\dfrac{\pi }{4}$.
Now, first we will calculate the value of the expression given in the question.
Now, first let us consider the L.H.S. = ${{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{3}{5}-{{\tan }^{-1}}\dfrac{8}{19}$
Now, we know that ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy},\text{ if xy1}$
Let us apply the formula on first two terms of the above equation we have
$\begin{align}
  & \left( {{\tan }^{-1}}\dfrac{3}{4}+{{\tan }^{-1}}\dfrac{3}{5} \right)-{{\tan }^{-1}}\dfrac{8}{19} \\
 & \Rightarrow {{\tan }^{-1}}\dfrac{\dfrac{3}{4}+\dfrac{3}{5}}{1-\dfrac{3}{4}\times \dfrac{3}{5}}-{{\tan }^{-1}}\dfrac{8}{19} \\
\end{align}$
Now, taking LCM and solving further we get
$\Rightarrow {{\tan }^{-1}}\dfrac{\dfrac{3\times 5+3\times 4}{4\times 5}}{1-\dfrac{9}{20}}-{{\tan }^{-1}}\dfrac{8}{19}$
Now, simplifying further we get
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}\dfrac{\dfrac{15+12}{20}}{\dfrac{20-9}{20}}-{{\tan }^{-1}}\dfrac{8}{19} \\
 & \Rightarrow {{\tan }^{-1}}\dfrac{\dfrac{27}{20}}{\dfrac{11}{20}}-{{\tan }^{-1}}\dfrac{8}{19} \\
\end{align}$
Or we can write that
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}\dfrac{27}{20}\times \dfrac{20}{11}-{{\tan }^{-1}}\dfrac{8}{19} \\
 & \Rightarrow {{\tan }^{-1}}\dfrac{27}{11}-{{\tan }^{-1}}\dfrac{8}{19} \\
\end{align}$
Now, we know that ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x-y}{1+xy},\text{ if xy-1}$
Now, applying the formula on the above equation we have
${{\tan }^{-1}}\dfrac{\dfrac{27}{11}-\dfrac{8}{19}}{1+\dfrac{27}{11}\times \dfrac{8}{19}}$
Now, taking LCM and solving further we get
$\Rightarrow {{\tan }^{-1}}\dfrac{\dfrac{27\times 19-8\times 11}{11\times 19}}{1+\dfrac{216}{209}}$
Now, simplifying further we get
$\begin{align}
  & \Rightarrow {{\tan }^{-1}}\dfrac{\dfrac{513-88}{209}}{\dfrac{209+216}{209}} \\
 & \Rightarrow {{\tan }^{-1}}\dfrac{\dfrac{425}{209}}{\dfrac{425}{209}} \\
\end{align}$
Or we can write that
$\Rightarrow {{\tan }^{-1}}\dfrac{425}{209}\times \dfrac{209}{425}$
So, we have $\Rightarrow {{\tan }^{-1}}1$
Now, we know that ${{\tan }^{-1}}1=\dfrac{\pi }{4}$ , which is equal to R.H.S.
L.H.S.=R.H.S.
Hence proved

Note: The inverse functions in the trigonometry are used to get the angle with any of the trigonometry ratio. Inverse trigonometric functions do the opposite of the regular trigonometric functions. We can also write ${{\tan }^{-1}}1$ as ${{\tan }^{-1}}1=A$ , so we have $\tan A=1$ and we know that tangent function gives the value $1$ when angle is equal to $\dfrac{\pi }{4}$ .
So, we have $\tan \dfrac{\pi }{4}=1$
Or we can write the above equation as\[{{\tan }^{-1}}1=\dfrac{\pi }{4}\] .