Question

Prove that \[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}\]

Answer 1

Hint: First expand the given expression in left hand side using the formula for expansion of \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\]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 L.H.S
\[{{\tan }^{-1}}\left( \dfrac{1}{3} \right)+{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{1}{8} \right)\]
As we can see we have to use \[{{\tan }^{-1}}x+{{\tan }^{-1}}y\]
Using the formula,
 \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Substituting \[x=\dfrac{1}{3}\]and \[y=\dfrac{1}{5}\]
Substituting \[x=\dfrac{1}{7}\]and \[y=\dfrac{1}{8}\]
\[={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{3}+\dfrac{1}{5}}{1-\left( \dfrac{1}{3} \right)\left( \dfrac{1}{5} \right)} \right)+{{\tan }^{-1}}\left( \dfrac{\dfrac{1}{7}+\dfrac{1}{8}}{1-\left( \dfrac{1}{7} \right)\left( \dfrac{1}{8} \right)} \right)\]
\[={{\tan }^{-1}}\left( \dfrac{\dfrac{5+3}{15}}{\dfrac{14}{15}} \right)+{{\tan }^{-1}}\left( \dfrac{\dfrac{8+7}{56}}{\dfrac{55}{56}} \right)\]
\[={{\tan }^{-1}}\left( \dfrac{8}{14} \right)+{{\tan }^{-1}}\left( \dfrac{15}{55} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Further solving (2) we get (3)
\[={{\tan }^{-1}}\left( \dfrac{8}{14} \right)+{{\tan }^{-1}}\left( \dfrac{3}{11} \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
\[={{\tan }^{-1}}\left( \dfrac{\dfrac{8}{14}+\dfrac{3}{11}}{1-\left( \dfrac{8}{14} \right)\left( \dfrac{3}{11} \right)} \right)\]
\[={{\tan }^{-1}}\left( \dfrac{\dfrac{130}{154}}{\dfrac{130}{154}} \right)\]
\[={{\tan }^{-1}}\left( 1 \right)\]
\[=\dfrac{\pi }{4}\]
= R.H.S

Note: If \[xy<1,{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]and if \[xy>1,{{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\], therefore it is always important to check the multiplication of $x$ and $y$ for every step , though here we did not do it because we could in the starting only that from initial only both $x$ and $y$ are less than 1 so their multiplication will always be less than 1.