Question

If x, y, z are in A.P. and ${\tan ^{ - 1}}x,{\tan ^{ - 1}}y,{\tan ^{ - 1}}z$ are also in A.P., then
A. 2x=3y=6z
B. 6x=3y=2z
C.6x=4y=3z
D. x=y=z

Answer 1

Hint: In order to solve this problem you need to know that the terms are in AP and at the same time they are in GP then the terms are equal. Here we will use the formula of tan2x and we will also use the formula of arithmetic mean and geometric mean to get the right answer.

Complete step-by-step answer:
We know that x, y, z are in AP so we can say that,
2y = x + z……………(1)(Arithmetic mean)
We also know that ${\tan ^{ - 1}}x,{\tan ^{ - 1}}y,{\tan ^{ - 1}}z$ are in AP so, we can say that,
$ \Rightarrow 2{\tan ^{ - 1}}y = {\tan ^{ - 1}}x + {\tan ^{ - 1}}z$(Arithmetic mean)
Using the formula ${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left[ {\dfrac{{a + b}}{{1 - ab}}} \right]$ we get the above equation as,
$ \Rightarrow 2{\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{x + z}}{{1 - xz}}} \right]$
On solving further we get,
$ \Rightarrow \tan \left( {2{{\tan }^{ - 1}}y} \right) = \dfrac{{x + z}}{{1 - xz}}$
Using the formula $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$ we get the above equation as,
$
   \Rightarrow \dfrac{{2\tan \left( {{{\tan }^{ - 1}}y} \right)}}{{1 - {{\left( {\tan \left( {{{\tan }^{ - 1}}y} \right)} \right)}^2}}} = \dfrac{{x + z}}{{1 - xz}} \\
   \Rightarrow \dfrac{{2y}}{{1 - {y^2}}} = \dfrac{{x + z}}{{1 - xz}} \\
$
From (1) we can say that $ \Rightarrow \dfrac{1}{{1 - {y^2}}} = \dfrac{1}{{1 - xz}}$
From the above equation we can clearly say that ${y^2} = xz$ and $x,y,z$ is a series so ${y^2} = xz$ is the geometric mean of the series it means $x,y,z$ are in GP also.
We know that the terms are in AP and at the same time they are in GP then the terms are equal.
So, x = y = z.
Therefore, the correct option is D.

Note: When you get to solve such problems just proceed solving with the help of given data at last you will get the result automatically during solving. Here we have used various formulas of trigonometry and inverse trigonometry mentioned above and mainly used the concept of arithmetic and geometric mean to get the right answer. Arithmetic is the average of a set of numerical values, as calculated by adding them together and dividing by the number of terms in the set. Geometric mean is the central number in a geometric progression (e.g. 9 in 3, 9, 27), also calculable as the nth root of a product of n numbers. Knowing this will help you to further solve your problem.