Question

Find the minimum value of $4{\tan }^2\theta +9{\cot }^2\theta$

• A. $6$
• B. $9$
• C. $12$
• D. $15$

Answer 1

The correct option is C $12$

Let $y=4{tan}^2\theta +9co{t}^2\theta$

$\therefore y=4{tan}^2\theta +\frac{9}{{tan}^2\theta }$

Put $tan\theta =x$

$\therefore \theta ={tan}^{-1}x$

Thus, $y=4x^2+\frac{9}{x^2}$

For y to be minimum, $\frac{dy}{dx}=0$

$\therefore \frac{d}{dx}[4x^2+\frac{9}{x^2}]=0$

$\therefore 8x-\frac{18}{x^3}=0$

$\therefore \frac{8x^4-18}{x^3}=0$

$\therefore 8x^4-18=0$

$\therefore 8x^4=18$

$\therefore x^4=\frac{18}{8}=\frac{9}{4}$

Taking square roots on both sides, we get,

$\therefore x^2=\frac{3}{2}$

Now, $\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\therefore \frac{d^{2}y}{{dx}^2}=\frac{d}{dx}(8x-\frac{18}{x^3})$

$\therefore \frac{d^{2}y}{{dx}^2}=8+\frac{54}{x^4}$

$\therefore \frac{d^{2}y}{{dx}^2}=8+\frac{54}{{\left(x^2\right)}^2}$

$\therefore \frac{d^{2}y}{{dx}^2}=8+\frac{54}{{\left(\frac{3}{2}\right)}^2}$

$\therefore \frac{d^{2}y}{{dx}^2}=8+\frac{54}{\frac{9}{4}}$

$\therefore \frac{d^{2}y}{{dx}^2}=8+(54\times \frac{4}{9})$

$\therefore \frac{d^{2}y}{{dx}^2}=8+24=32$

$\therefore \frac{d^{2}y}{{dx}^2}>0$

Thus,

$y$ is minimum at $x^2=\frac{3}{2}$

Thus,

$y_{min}=4\left(\frac{3}{2}\right)+\frac{9}{\left(\frac{3}{2}\right)}$

$y_{min}=6+(9\times \frac{2}{3})$

$y_{min}=6+6$

$y_{min}=12$