Question

If the non-zero numbers a, b, c are in A.P. and ${\tan ^{ - 1}}a$ , ${\tan ^{ - 1}}b$ and ${\tan ^{ - 1}}c$ are also in A.P., then prove that \[a = b = c\] and ${b^2} = ac$ .

Answer 1

Hint: If x, y, z are in A.P., then \[2x = y + z\] .
Using the same property as mentioned above, find the relation between a, b, c and ${\tan ^{ - 1}}a$ , ${\tan ^{ - 1}}b$ , ${\tan ^{ - 1}}c$ .
Now, using ${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$ , and solving it further, we will get to prove the relation ${b^2} = ac$ .
Now, using ${b^2} = ac$ , prove \[a = b = c\] .

Complete step-by-step answer:
It is given that the non-zero numbers a, b, c are in A.P.
 \[\therefore \;2b = a + c\] ... (1)
Also, ${\tan ^{ - 1}}a$ , ${\tan ^{ - 1}}b$ and ${\tan ^{ - 1}}c$ are also in A.P.
 $\therefore 2{\tan ^{ - 1}}b = {\tan ^{ - 1}}a + {\tan ^{ - 1}}c$
Now, substituting $2{\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{2b}}{{1 - {b^2}}}} \right)$ and ${\tan ^{ - 1}}a + {\tan ^{ - 1}}c = {\tan ^{ - 1}}\left( {\dfrac{{a + c}}{{1 - ac}}} \right)$ .
 $
  \therefore {\tan ^{ - 1}}\left( {\dfrac{{2b}}{{1 - {b^2}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{a + c}}{{1 - ac}}} \right) \\
  \therefore \dfrac{{2b}}{{1 - {b^2}}} = \dfrac{{a + c}}{{1 - ac}} \\
 $
Substitute \[2b = a + c\] , from equation (1)
 $
  \therefore \dfrac{{a + c}}{{1 - {b^2}}} = \dfrac{{a + c}}{{1 - ac}} \\
  \therefore \dfrac{1}{{1 - {b^2}}} = \dfrac{1}{{1 - ac}} \\
  \therefore 1 - ac = 1 - {b^2} \\
  \therefore {b^2} = 1 - 1 + ac \\
 $
 $\therefore {b^2} = ac$ (Proved)
Now, to prove \[a = b = c\] , we have to take help of ${b^2} = ac$ .
 $
  \therefore {b^2} = ac \\
  \therefore 4{b^2} = 4ac \\
  \therefore {\left( {2b} \right)^2} - 4ac = 0 \\
 $
Substitute \[2b = a + c\] from equation (1)
 $
  \therefore {\left( {a + c} \right)^2} - 4ac = 0 \\
  \therefore {a^2} + 2ac + {c^2} - 4ac = 0 \\
  \therefore {a^2} - 2ac + {c^2} = 0 \\
  \therefore {\left( {a - c} \right)^2} = 0 \\
  \therefore a - c = 0 \\
  \therefore a = c \\
 $
We have \[a = c\] ... (2)
Now, substitute \[a = c\] in ${b^2} = ac$ .
 $
  \therefore {b^2} = a\left( a \right) \\
  \therefore {b^2} = {a^2} \\
  \therefore b = a \\
 $
We now have, \[a = b\] and also \[a = c\] .
Thus, \[a = b = c\] . (Proved)

Note: Here, to prove \[a = b = c\] , we have to take help of ${b^2} = ac$ by multiplying the whole equation by 4.
Also, we know that ${\left( {a + c} \right)^2} - 4ac = 0$ can directly be written as ${\left( {a - c} \right)^2} = 0$ by using property ${\left( {x + y} \right)^2} - 4xy = {\left( {x - y} \right)^2}$ . So, to get the answer quickly, we can directly write ${\left( {a + c} \right)^2} - 4ac = 0$ as ${\left( {a - c} \right)^2} = 0$ and not solve ${\left( {a + c} \right)^2} - 4ac = 0$ to get ${\left( {a - c} \right)^2} = 0$ .