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Hint: In this question we have to find the value of x + y + z, make use of basic trigonometric identity of ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$ to simplify the left hand side of the given equation with respect to the right hand side. This concept will help in getting the right answer.

Complete step-by-step answer:

Given equation is

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi $

Then we have to find out the value of $x + y + z$.

Now as we know that ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$ so, use this property in above equation we have,

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

Now again apply the property

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

Now simplify the above equation we have,

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

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

Now shift tan inverse to R.H.S

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

This condition only holds when the denominator of L.H.S is not zero.

Therefore the denominator of L.H.S should be less than zero or greater than zero.

But it is given that $xy + xz + yz < 1$………………. (2)

Therefore the denominator of L.H.S is not zero according to equation (2)

So, equation (1) holds.

Now as we know that the value of $\tan \pi $ is zero. So, substitute this value in above equation we have,

$ \Rightarrow \left( {\dfrac{{x + y + z - xyz}}{{1 - xy - xz - yz}}} \right) = 0$

$ \Rightarrow x + y + z - xyz = 0$

$ \Rightarrow x + y + z = xyz$

So, xyz is the required answer of $x + y + z$.

Hence, option (b) is correct.

Note: Whenever we face such types of problems the key point is simply to have a good grasp of the inverse trigonometric identities some of them are mentioned above. The application of these identities will help you get on the right track to reach the solution.

Complete step-by-step answer:

Given equation is

${\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi $

Then we have to find out the value of $x + y + z$.

Now as we know that ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$ so, use this property in above equation we have,

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

Now again apply the property

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

Now simplify the above equation we have,

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

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

Now shift tan inverse to R.H.S

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

This condition only holds when the denominator of L.H.S is not zero.

Therefore the denominator of L.H.S should be less than zero or greater than zero.

But it is given that $xy + xz + yz < 1$………………. (2)

Therefore the denominator of L.H.S is not zero according to equation (2)

So, equation (1) holds.

Now as we know that the value of $\tan \pi $ is zero. So, substitute this value in above equation we have,

$ \Rightarrow \left( {\dfrac{{x + y + z - xyz}}{{1 - xy - xz - yz}}} \right) = 0$

$ \Rightarrow x + y + z - xyz = 0$

$ \Rightarrow x + y + z = xyz$

So, xyz is the required answer of $x + y + z$.

Hence, option (b) is correct.

Note: Whenever we face such types of problems the key point is simply to have a good grasp of the inverse trigonometric identities some of them are mentioned above. The application of these identities will help you get on the right track to reach the solution.

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