Solveeit Logo
Login

Question

Question: : Prove that \({{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 ...

: Prove that tan⁡−1(1)+tan⁡−1(2)+tan⁡−1(3)=π{{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 \right)=\pi .

Explanation

Solution

Hint: Assume three variables (say x, y, z) for tan⁡−1(1),tan⁡−12 andtan⁡−13{{\tan }^{-1}}\left( 1 \right),{{\tan }^{-1}}2\ and {{\tan }^{-1}}3 respectively. Then use the formula,
′′tan⁡(x+y+z)=tan⁡x+tan⁡y+tan⁡z−tan⁡xtan⁡ytan⁡z1−tan⁡xtan⁡y−tan⁡z.tan⁡y−tan⁡ytan⁡z′′''\tan \left( x+y+z \right)=\dfrac{\tan x+\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan z.\tan y-\tan y\tan z}''.
Put the value of tan⁡x,tan⁡y and tan⁡z\tan x,\tan y\ and\ \tan z to get the value of tan⁡(x+y+z)\tan \left( x+y+z \right) and then with the help of tan⁡(x+y+z)\tan \left( x+y+z \right), find the value of x+y+zx+y+z.

Complete Step-by-step answer:
To prove: tan⁡−1(1)+tan⁡−1(2)+tan⁡−1(3)=π{{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 \right)=\pi
Proof:
LHS=tan⁡−1(1)+tan⁡−1(2)+tan⁡−1(3) RHS=π \begin{aligned} & LHS={{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 \right) \\\ & RHS=\pi \\\ \end{aligned}
We have to prove LHS = RHS
Let us start with LHS.
LHS=tan⁡−1(1)+tan⁡−1(2)+tan⁡−1(3)LHS={{\tan }^{-1}}\left( 1 \right)+{{\tan }^{-1}}\left( 2 \right)+{{\tan }^{-1}}\left( 3 \right)
Let us assume x=tan⁡−11x={{\tan }^{-1}}1
Taking tan both sides, we will get,
tan⁡(x)=tan⁡(tan⁡−11) ⇒tan⁡x=1 \begin{aligned} & \tan \left( x \right)=\tan \left( {{\tan }^{-1}}1 \right) \\\ & \Rightarrow \tan x=1 \\\ \end{aligned}
Similarly, let us assume y=tan⁡−12y={{\tan }^{-1}}2
Taking tan both sides, we will get,
tan⁡y=tan⁡(tan⁡−12) ⇒tan⁡y=2 \begin{aligned} & \tan y=\tan \left( {{\tan }^{-1}}2 \right) \\\ & \Rightarrow \tan y=2 \\\ \end{aligned}
Similarly, let us assume z=tan⁡−13z={{\tan }^{-1}}3
Taking tan both sides, we will get,
tan⁡z=tan⁡(tan⁡−13) ⇒tan⁡z=3 \begin{aligned} & \tan z=\tan \left( {{\tan }^{-1}}3 \right) \\\ & \Rightarrow \tan z=3 \\\ \end{aligned}
Now, let us use the formula,
tan⁡(x+y+z)=tan⁡x+tan⁡y+tan⁡z−tan⁡xtan⁡ytan⁡z1−tan⁡xtan⁡y−tan⁡z.tan⁡y−tan⁡ytan⁡z\tan \left( x+y+z \right)=\dfrac{\tan x+\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan z.\tan y-\tan y\tan z}
On putting tan x = 1, tan y = 2 and tan z = 3 as calculated above, we will get,

⇒tan⁡(x+y+z)=1+2+3−(1)(2)(3)1−(1)(2)−(1)(3)−(2)(3) ⇒tan⁡(x+y+z)=6−61−2−3−6 ⇒tan⁡(x+y+z)=0−10 ⇒tan⁡(x+y+z)=0 \begin{aligned} & \Rightarrow \tan \left( x+y+z \right)=\dfrac{1+2+3-\left( 1 \right)\left( 2 \right)\left( 3 \right)}{1-\left( 1 \right)\left( 2 \right)-\left( 1 \right)\left( 3 \right)-\left( 2 \right)\left( 3 \right)} \\\ & \Rightarrow \tan \left( x+y+z \right)=\dfrac{6-6}{1-2-3-6} \\\ & \Rightarrow \tan \left( x+y+z \right)=\dfrac{0}{-10} \\\ & \Rightarrow \tan \left( x+y+z \right)=0 \\\ \end{aligned}
We know,
tan⁡π=0 ⇒x+y+z=π \begin{aligned} & \tan \pi =0 \\\ & \Rightarrow x+y+z=\pi \\\ \end{aligned}
Replace x with tan⁡−11 and y with tan⁡−12 and z with tan⁡−13x\ with\ {{\tan }^{-1}}1\ and\ y\ with\ {{\tan }^{-1}}2\ and\ z\ with\ {{\tan }^{-1}}3, we will get,
tan⁡−11+tan⁡−12+tan⁡−13=π{{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3=\pi
Proved.

Note: Note that tan (0) is also equal to 0. But tan⁡−11+tan⁡−12+tan⁡−13{{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3 can’t be equal to zero. As tan of these three angles are positive which mean all these angles are greater than zero and thus their sum can’t be zero. Students can make mistakes in the last step by taking x+y+z=0.