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Question: Prove that \[{{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{3}{5} \right)-{{\...

Prove that tan⁡−1(34)+tan⁡−1(35)−tan⁡−1(819)=π4{{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{3}{5} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)=\dfrac{\pi }{4}

Explanation

Solution

First expand the given expression in left hand side using the formula for expansion of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y and tan⁡−1x−tan⁡−1y{{\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(34)+tan⁡−1(35)−tan⁡−1(819){{\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⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y
We know that
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Substituting x=34x=\dfrac{3}{4}, y=35y=\dfrac{3}{5}
=tan⁡−1(34+351−(34)(35))−tan⁡−1(819)={{\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(15+122020−920)−tan⁡−1(819)={{\tan }^{-1}}\left( \dfrac{\dfrac{15+12}{20}}{\dfrac{20-9}{20}} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)
=tan⁡−1(27201120)−tan⁡−1(819)={{\tan }^{-1}}\left( \dfrac{\dfrac{27}{20}}{\dfrac{11}{20}} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right)
=tan⁡−1(2711)−tan⁡−1(819)={{\tan }^{-1}}\left( \dfrac{27}{11} \right)-{{\tan }^{-1}}\left( \dfrac{8}{19} \right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
The above expression (a) is in the form tan⁡−1x−tan⁡−1y{{\tan }^{-1}}x-{{\tan }^{-1}}y
We know that tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy){{\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((2711)−(819)1+(2711)(819))={{\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(513−88209209+21620)={{\tan }^{-1}}\left( \dfrac{\dfrac{513-88}{209}}{\dfrac{209+216}{20}} \right)
=tan⁡−1(425209425209)={{\tan }^{-1}}\left( \dfrac{\dfrac{425}{209}}{\dfrac{425}{209}} \right)
=tan⁡−1(1)={{\tan }^{-1}}\left( 1 \right)
= R.H.S

Note: Usage of the formulas (tan⁡−1x+tan⁡−1y)\left( {{\tan }^{-1}}x+{{\tan }^{-1}}y \right) and (tan⁡−1x−tan⁡−1y)\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.