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

Answer

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Hint: First of all, we can use the trigonometric relation, given by \[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]. Then, we can apply the relation again to the previous results and \[{{\tan }^{-1}}z\]. 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 \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \]. We have to prove that \[x+y+z=xyz\].

Let us consider the equation given in the question, that is,

\[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi ....\left( i \right)\]

We know that, \[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\]. By substituting A = x and B = y, we get,

\[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]

By substituting the value of \[\left( {{\tan }^{-1}}x+{{\tan }^{-1}}y \right)\] in equation (i), we get,

\[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}z=\pi ....\left( ii \right)\]

Again, we know that \[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\]

By substituting \[A=\dfrac{\left( x+y \right)}{\left( 1-xy \right)}\] and B = z, we get,

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}\left( z \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{x+y}{1-xy} \right)+z}{1-\left( \dfrac{x+y}{1-xy} \right)z} \right)\]

By substituting the value of \[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}\left( z \right)=\pi \] from equation (ii) in the above equation, we get,

\[\Rightarrow \pi ={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{x+y}{1-xy} \right)+z\left( 1-xy \right)}{\dfrac{\left( 1-xy \right)-\left( x+y \right)z}{\left( 1-xy \right)}} \right)\]

By canceling the like terms in the RHS of the above equation, we get,

\[\Rightarrow \pi ={{\tan }^{-1}}\left[ \dfrac{\left( x+y \right)+\left( z-xyz \right)}{1-xy-\left( xz+yz \right)} \right]\]

By taking tan on both sides of the above equation, we get,

\[\tan \pi =\tan \left( {{\tan }^{-1}}\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right] \right)\]

We know that \[\tan \pi =0\]. By substituting this in the above equation, we get,

\[\tan {{\tan }^{-1}}\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right]=0\]

Also, we know that \[\tan {{\tan }^{-1}}A=A\]. By using this in the above equation, we get,

\[\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right]=0\]

By cross multiplying above equation, we get,

\[x+y+z-xyz=0\]

By adding xyz on both sides of the above equation, we get,

\[x+y+z=xyz\]

Hence proved.

Note: Students can also solve this question by taking tan on both sides in the first step only and using the formula \[\tan \left( A+B+C \right)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\] by taking \[{{\tan }^{-1}}x=A,{{\tan }^{-1}}y=B\text{ and }{{\tan }^{-1}}z=C\] and then using \[\tan {{\tan }^{-1}}A=A\] 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.

Complete step-by-step answer:

Here, we are given that \[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi \]. We have to prove that \[x+y+z=xyz\].

Let us consider the equation given in the question, that is,

\[{{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z=\pi ....\left( i \right)\]

We know that, \[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\]. By substituting A = x and B = y, we get,

\[{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\]

By substituting the value of \[\left( {{\tan }^{-1}}x+{{\tan }^{-1}}y \right)\] in equation (i), we get,

\[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}z=\pi ....\left( ii \right)\]

Again, we know that \[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\]

By substituting \[A=\dfrac{\left( x+y \right)}{\left( 1-xy \right)}\] and B = z, we get,

\[\Rightarrow {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}\left( z \right)={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{x+y}{1-xy} \right)+z}{1-\left( \dfrac{x+y}{1-xy} \right)z} \right)\]

By substituting the value of \[{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)+{{\tan }^{-1}}\left( z \right)=\pi \] from equation (ii) in the above equation, we get,

\[\Rightarrow \pi ={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{x+y}{1-xy} \right)+z\left( 1-xy \right)}{\dfrac{\left( 1-xy \right)-\left( x+y \right)z}{\left( 1-xy \right)}} \right)\]

By canceling the like terms in the RHS of the above equation, we get,

\[\Rightarrow \pi ={{\tan }^{-1}}\left[ \dfrac{\left( x+y \right)+\left( z-xyz \right)}{1-xy-\left( xz+yz \right)} \right]\]

By taking tan on both sides of the above equation, we get,

\[\tan \pi =\tan \left( {{\tan }^{-1}}\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right] \right)\]

We know that \[\tan \pi =0\]. By substituting this in the above equation, we get,

\[\tan {{\tan }^{-1}}\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right]=0\]

Also, we know that \[\tan {{\tan }^{-1}}A=A\]. By using this in the above equation, we get,

\[\left[ \dfrac{x+y+z-xyz}{1-\left( xy+yz+zx \right)} \right]=0\]

By cross multiplying above equation, we get,

\[x+y+z-xyz=0\]

By adding xyz on both sides of the above equation, we get,

\[x+y+z=xyz\]

Hence proved.

Note: Students can also solve this question by taking tan on both sides in the first step only and using the formula \[\tan \left( A+B+C \right)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C-\tan C\tan A}\] by taking \[{{\tan }^{-1}}x=A,{{\tan }^{-1}}y=B\text{ and }{{\tan }^{-1}}z=C\] and then using \[\tan {{\tan }^{-1}}A=A\] 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.

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