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Question: If \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \]. Then prove that \[x+y+z=xyz\]....

If tan⁡−1x+tan⁡−1y+tan⁡−1z=π{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi . Then prove that x+y+z=xyzx+y+z=xyz.

Explanation

Solution

Hint: First of all, we can use the trigonometric relation, given by tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right). Then, we can apply the relation again to the previous results and tan⁡−1z{{\tan }^{-1}}z. Then take tan on both sides, and then solve the equation to get the desired result.

Complete step-by-step answer:
Here, we are given that tan⁡−1x+tan⁡−1y+tan⁡−1z=π{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi . We have to prove that x+y+z=xyzx+y+z=xyz.
Let us consider the equation given in the question, that is,
tan⁡−1x+tan⁡−1y+tan⁡−1z=π....(i){{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi ....\left( i \right)
We know that, tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right). By substituting A = x and B = y, we get,
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)
By substituting the value of (tan⁡−1x+tan⁡−1y)\left( {{\tan }^{-1}}x+{{\tan }^{-1}}y \right) in equation (i), we get,
tan⁡−1(x+y1−xy)+tan⁡−1z=π....(ii){{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}z=\pi ....\left( ii \right)
Again, we know that tan⁡−1A+tan⁡−1B=tan⁡−1(A+B1−AB){{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)
By substituting A=(x+y)(1−xy)A=\dfrac{\left( x+y \right)}{\left( 1-xy \right)} and B = z, we get,
⇒tan⁡−1(x+y1−xy)+tan⁡−1(z)=tan⁡−1((x+y1−xy)+z1−(x+y1−xy)z)\Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}\left( z \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{x+y}{1-xy} \right)+z}{1-\left( \dfrac{x+y}{1-xy} \right)z} \right)
By substituting the value of tan⁡−1(x+y1−xy)+tan⁡−1(z)=π{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}\left( z \right)=\pi from equation (ii) in the above equation, we get,
⇒π=tan⁡−1((x+y1−xy)+z(1−xy)(1−xy)−(x+y)z(1−xy))\Rightarrow \pi ={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{x+y}{1-xy} \right)+z\left( 1-xy \right)}{\dfrac{\left( 1-xy \right)-\left( x+y \right)z}{\left( 1-xy \right)}} \right)
By canceling the like terms in the RHS of the above equation, we get,
⇒π=tan⁡−1[(x+y)+(z−xyz)1−xy−(xz+yz)]\Rightarrow \pi ={{\tan }^{-1}}\left[ \dfrac{\left( x+y \right)+\left( z-xyz \right)}{1-xy-\left( xz+yz \right)} \right]
By taking tan on both sides of the above equation, we get,
tan⁡π=tan⁡(tan⁡−1[x+y+z−xyz1−(xy+yz+zx)])\tan \pi =\tan \left( {{\tan }^{-1}}\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right] \right)
We know that tan⁡π=0\tan \pi =0. By substituting this in the above equation, we get,
tan⁡tan⁡−1[x+y+z−xyz1−(xy+yz+zx)]=0\tan {{\tan }^{-1}}\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right]=0
Also, we know that tan⁡tan⁡−1A=A\tan {{\tan }^{-1}}A=A. By using this in the above equation, we get,
[x+y+z−xyz1−(xy+yz+zx)]=0\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right]=0
By cross multiplying above equation, we get,
x+y+z−xyz=0x+y+z-xyz=0
By adding xyz on both sides of the above equation, we get,
x+y+z=xyzx+y+z=xyz
Hence proved.

Note: Students can also solve this question by taking tan on both sides in the first step only and using the formula tan⁡(A+B+C)=tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A\tan \left( A+B+C \right)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A} by taking tan⁡−1x=A,tan⁡−1y=B and tan⁡−1z=C{{\tan }^{-1}}x=A,{{\tan }^{-1}}y=B\text{ and }{{\tan }^{-1}}z=C and then using tan⁡tan⁡−1A=A\tan {{\tan }^{-1}}A=A to get the desired result. Memorizing this formula is easy but it gets complicated while applying the values and simplifying to get the result. The chances of making silly mistakes can also arise.