Question

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

Answer 1

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.

Complete step by step answer:
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.