Question

If ‘a’ is a real constant and A,B,C are the variable angles and \[\sqrt{{{a}^{2}}-4}\tan A+a\tan B+\sqrt{{{a}^{2}}+4}\tan C=6a\] , then the least value of \[{{\tan }^{2}}A+{{\tan }^{2}}B+{{\tan }^{2}}C\] is
\[\begin{align}
  & (A)6 \\
 & (B)10 \\
 & (C)12 \\
 & (D)3 \\
\end{align}\]

Answer 1

Hint: We should know that that if \[{{a}_{1}},{{a}_{2}},{{a}_{3}},{{b}_{1}},{{b}_{2}},{{b}_{3}}\] are numbers, integers, functions etc….. then \[{{({{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}})}^{2}}\ge \left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)\]. Now we should compare \[\sqrt{{{a}^{2}}-4}\tan A+a\tan B+\sqrt{{{a}^{2}}+4}\tan C\] with \[{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}\]. Now we can get the values of \[{{a}_{1}},{{a}_{2}},{{a}_{3}},{{b}_{1}},{{b}_{2}},{{b}_{3}}\].
Now by using the statement \[{{({{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}})}^{2}}\ge \left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)\], we can find the values of \[{{\tan }^{2}}A+{{\tan }^{2}}B+{{\tan }^{2}}C\].

Complete step by step solution:
Before solving the problem, we should know that if \[{{a}_{1}},{{a}_{2}},{{a}_{3}},{{b}_{1}},{{b}_{2}},{{b}_{3}}\] are numbers, integers, functions etc….. then \[{{({{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}})}^{2}}\ge \left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)\].
From the question, we given that \[\sqrt{{{a}^{2}}-4}\tan A+a\tan B+\sqrt{{{a}^{2}}+4}\tan C=6a\]
Let us compare \[\sqrt{{{a}^{2}}-4}\tan A+a\tan B+\sqrt{{{a}^{2}}+4}\tan C\] with \[{{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}}\], then we get
\[\begin{align}
  & {{a}_{1}}=\sqrt{{{a}^{2}}+4}....(1) \\
 & {{a}_{2}}=a.....(2) \\
 & {{a}_{3}}=\sqrt{{{a}^{2}}+4}....(3) \\
 & {{b}_{1}}=\tan A.....(4) \\
 & {{b}_{2}}=\tan B.....(5) \\
 & {{b}_{3}}=\tan C.....(6) \\
\end{align}\]
Let us assume
\[\sqrt{{{a}^{2}}-4}\tan A+a\tan B+\sqrt{{{a}^{2}}+4}\tan C=6a......(7)\]
From (1), (2), (3), (4), (5), (6) and (7)
Now we will apply the condition for the given equation.
\[{{({{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}})}^{2}}\ge \left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)\]
\[\Rightarrow {{\left( \sqrt{{{a}^{2}}-4}\tan A+a\tan B+\sqrt{{{a}^{2}}+4}\tan C \right)}^{2}}\ge \left( {{\left( \sqrt{{{a}^{2}}-4} \right)}^{2}}+{{a}^{2}}+{{\left( \sqrt{{{a}^{2}}+4} \right)}^{2}} \right)\left( ta{{n}^{2}}A+ta{{n}^{2}}B+ta{{n}^{2}}C \right)\]
\[\Rightarrow {{\left( \sqrt{{{a}^{2}}-4}\tan A+a\tan B+\sqrt{{{a}^{2}}+4}\tan C \right)}^{2}}\ge \left( 3{{a}^{2}} \right)\left( ta{{n}^{2}}A+ta{{n}^{2}}B+ta{{n}^{2}}C \right)......(8)\]
Now we will substitute equation (7) in equation (8).
\[\begin{align}
  & \Rightarrow {{\left( 6a \right)}^{2}}\ge \left( 3{{a}^{2}} \right)\left( ta{{n}^{2}}A+ta{{n}^{2}}B+ta{{n}^{2}}C \right) \\
 & \Rightarrow 36{{a}^{2}}\ge (3{{a}^{2}})\left( ta{{n}^{2}}A+ta{{n}^{2}}B+ta{{n}^{2}}C \right) \\
 & \Rightarrow \left( ta{{n}^{2}}A+ta{{n}^{2}}B+ta{{n}^{2}}C \right)\le 12 \\
\end{align}\]
So, the least value of \[{{\tan }^{2}}A+{{\tan }^{2}}B+{{\tan }^{2}}C\]is equal to 12.
Hence, option C is correct.

Note: Some students have a misconception that if \[{{a}_{1}},{{a}_{2}},{{a}_{3}},{{b}_{1}},{{b}_{2}},{{b}_{3}}\] are numbers, integers, functions etc….. then \[{{({{a}_{1}}{{b}_{1}}+{{a}_{2}}{{b}_{2}}+{{a}_{3}}{{b}_{3}})}^{2}}\le \left( a_{1}^{2}+a_{2}^{2}+a_{3}^{2} \right)\left( b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \right)\]. If this misconception is followed, we will get the maximum value of \[{{\tan }^{2}}A+{{\tan }^{2}}B+{{\tan }^{2}}C\] is equal to 12. But we want the minimum value of \[{{\tan }^{2}}A+{{\tan }^{2}}B+{{\tan }^{2}}C\]. The minimum cannot be obtained. So, to solve this problem students should have a clear view of the concept of inequalities.