Solveeit Logo
Login

Question

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

Prove that tan⁡−1(13)+tan⁡−1(15)+tan⁡−1(17)+tan⁡−1(18)=π4{{\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}

Explanation

Solution

Hint: First expand the given expression in left hand side using the formula for expansion of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}ynow 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(13)+tan⁡−1(15)+tan⁡−1(17)+tan⁡−1(18){{\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⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y
Using the formula,
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=13x=\dfrac{1}{3}and y=15y=\dfrac{1}{5}
Substituting x=17x=\dfrac{1}{7}and y=18y=\dfrac{1}{8}
=tan⁡−1(13+151−(13)(15))+tan⁡−1(17+181−(17)(18))={{\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(5+3151415)+tan⁡−1(8+7565556)={{\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(814)+tan⁡−1(1555)={{\tan }^{-1}}\left( \dfrac{8}{14} \right)+{{\tan }^{-1}}\left( \dfrac{15}{55} \right). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Further solving (2) we get (3)
=tan⁡−1(814)+tan⁡−1(311)={{\tan }^{-1}}\left( \dfrac{8}{14} \right)+{{\tan }^{-1}}\left( \dfrac{3}{11} \right) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
=tan⁡−1(814+3111−(814)(311))={{\tan }^{-1}}\left( \dfrac{\dfrac{8}{14}+\dfrac{3}{11}}{1-\left( \dfrac{8}{14} \right)\left( \dfrac{3}{11} \right)} \right)
=tan⁡−1(130154130154)={{\tan }^{-1}}\left( \dfrac{\dfrac{130}{154}}{\dfrac{130}{154}} \right)
=tan⁡−1(1)={{\tan }^{-1}}\left( 1 \right)
=π4=\dfrac{\pi }{4}
= R.H.S

Note: If xy<1,tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)xy<1,{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)and if xy>1,tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy)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 xx and yy for every step , though here we did not do it because we could in the starting only that from initial only both xx and yy are less than 1 so their multiplication will always be less than 1.