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

& (A)6 \\\ & (B)10 \\\ & (C)12 \\\ & (D)3 \\\ \end{aligned}$$

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

& {{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{aligned}$$ Let us assume $$\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{aligned} & \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{aligned}$$ So, the least value of $${{\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.