Question

The minimum value of $9{\tan ^2}\theta + 4{\cot ^2}\theta $ is:
A) $13$
B) $9$
C) $6$
D) $12$

Answer 1

Hint: The minimum value of trigonometric ratios and their sums are based on a multitude of factors and there is no one “fit all” way to find the minimum value of trigonometric function. A fair bit of intuition is also involved in these types of solutions. We will first convert the expression into a perfect square equation for $\tan $& $\cot $. Then solve it further to get the minimum value.

Complete step by step answer:
We will first try to make a perfect square for $\tan $& $\cot $. We will proceed as follows:
       \[9{\text{ }}ta{n^2}\theta {\text{ }} + {\text{ }}4{\text{ }}co{t^2}\theta \] (given)
\[ = {\text{ }}{\left( {3{\text{ }}tan{\text{ }}\theta } \right)^2}\; + {\text{ }}{\left( {2{\text{ }}cot{\text{ }}\theta } \right)^2}\]
\[ = {\text{ }}[{\left( {3{\text{ }}tan{\text{ }}\theta } \right)^2}\; + {\text{ }}{\left( {2{\text{ }}cot{\text{ }}\theta } \right)^2}\;-{\text{ }}12]{\text{ }} + {\text{ }}12\]
Now we the first part of the expression which is under the bracket is a perfect square because we can write the first term as : \[{\left( {3{\text{ }}tan{\text{ }}\theta {\text{ }}-{\text{ }}2{\text{ }}cot{\text{ }}\theta } \right)^2}\]
Thus writing it that way we get,
\[9{\text{ }}ta{n^2}\theta {\text{ }} + {\text{ }}4{\text{ }}co{t^2}\theta = {\text{ }}{\left( {3{\text{ }}tan{\text{ }}\theta {\text{ }}-{\text{ }}2{\text{ }}cot{\text{ }}\theta } \right)^2}\; + {\text{ }}12\]
Now for this expression to be minimum the whole square part should be minimum, now to make that whole square part minimum we put it equal to zero as square cannot be negative so the zero will be the minimum,
\[3{\text{ }}tan{\text{ }}\theta {\text{ }}-{\text{ }}2{\text{ }}cot{\text{ }}\theta {\text{ }} = {\text{ }}0\]
Since we have to only find the minimum value of the expression not the $\theta $ at which the value is minimum we can write our expression as :
\[9{\text{ }}ta{n^2}\theta {\text{ }} + {\text{ }}4{\text{ }}co{t^2}\theta = {\text{ }}{0^2}\; + {\text{ }}12\]
\[9{\text{ }}ta{n^2}\theta {\text{ }} + {\text{ }}4{\text{ }}co{t^2}\theta = {\text{ }}12\].
Thus $12$ will be the minimum value of the expression \[9{\text{ }}ta{n^2}\theta {\text{ }} + {\text{ }}4{\text{ }}co{t^2}\theta \] which is given as the option (D).

Note:
In case the question asked the value at which the expression \[9{\text{ }}ta{n^2}\theta {\text{ }} + {\text{ }}4{\text{ }}co{t^2}\theta \] becomes minimum we would have solved the equation :
\[3{\text{ }}tan{\text{ }}\theta {\text{ }}-{\text{ }}2{\text{ }}cot{\text{ }}\theta {\text{ }} = {\text{ }}0\]
That we had skipped , it would have been solved as,
\[3{\text{ }}tan{\text{ }}\theta {\text{ }}-{\text{ }}2{\text{ }}cot{\text{ }}\theta {\text{ }} = {\text{ }}0\]
\[3{\text{ }}tan{\text{ }}\theta {\text{ = }}2{\text{ }}cot{\text{ }}\theta {\text{ }}\]
\[{\text{ }}ta{n^2}{\text{ }}\theta {\text{ }} = {\text{ }}\dfrac{2}{3}\]
Taking roots of the both sides we get,
\[{\text{ }}tan{\text{ }}\theta {\text{ = }}\sqrt {\dfrac{2}{3}} \]
$\theta = ta{n^{ - 1}}(1.73)$