Courses
Courses for Kids
Free study material
Offline Centres
More

# Prove that ${{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)=\dfrac{\pi }{4}$

Last updated date: 26th Feb 2024
Total views: 360.3k
Views today: 4.60k
Verified
360.3k+ views
Hint: First expand the given expression in left hand side using the formula for expansion of ${{\tan }^{-1}}x+{{\tan }^{-1}}y$ and ${{\tan }^{-1}}x-{{\tan }^{-1}}y$ now substitute the values of x , y according to given expression and do the basic mathematical operations like addition and multiplication to get the required expression in the right hand side.

Now considering the L.H.S
L.H.S = ${{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)$
The first two terms are in the form of ${{\tan }^{-1}}x+{{\tan }^{-1}}y$
We know that
${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Substituting $x=\dfrac{3}{4}$, $y=\dfrac{3}{5}$
$={{\tan }^{-1}}\left( \dfrac{\dfrac{3}{4}+\dfrac{3}{5}}{1-\left( \dfrac{3}{4} \right)\left( \dfrac{3}{5} \right)} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)$
$={{\tan }^{-1}}\left( \dfrac{\dfrac{15+12}{20}}{\dfrac{20-9}{20}} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)$
$={{\tan }^{-1}}\left( \dfrac{\dfrac{27}{20}}{\dfrac{11}{20}} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)$
$={{\tan }^{-1}}\left( \dfrac{27}{11} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
The above expression (a) is in the form ${{\tan }^{-1}}x-{{\tan }^{-1}}y$
We know that ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substituting the values of x and y in (2) we get,
$={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{27}{11} \right)-\left( \dfrac{8}{19} \right)}{1+\left( \dfrac{27}{11} \right)\left( \dfrac{8}{19} \right)} \right)$
$={{\tan }^{-1}}\left( \dfrac{\dfrac{513-88}{209}}{\dfrac{209+216}{20}} \right)$
$={{\tan }^{-1}}\left( \dfrac{\dfrac{425}{209}}{\dfrac{425}{209}} \right)$
$={{\tan }^{-1}}\left( 1 \right)$
= R.H.S

Note: Usage of the formulas $\left( {{\tan }^{-1}}x+{{\tan }^{-1}}y \right)$ and $\left( {{\tan }^{-1}}x-{{\tan }^{-1}}y \right)$ should be done carefully to simplify the given question and application of the formulas in correct way is necessary.