Question

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

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

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