Minimizing $4\sec^{2}(x)+9\csc^{2}(x)$ for $x$ in the first quadrant. Discrepancy in solution

Question

Given that $x$ lies in the first quadrant. Find the minimum value of $$4\sec^{2}(x)+9\csc^{2}(x)$$

Using derivatives, I am able to show that the minimum value of the expression-in-question is equal to $25$. I also verified this with Desmos graphing app. However, when I tried doing this using basic algebra, the answer turns out to be $26$. I do not know why this is happening, but I should be extremely grateful to you if you can point out the error.

$$4\sec^{2}(x)+9\csc^{2}(x)=$$ $$(2\sec(x)-3\csc(x))^{2}+12\sec(x)\csc(x)$$

The value of the above expression will be minimum when that expression inside parenthesis equals zero; and that happens when $\tan(x)=\frac{3}{2}$. Using this we can say that $\sec(x)=\frac{\sqrt{13}}{2}$ and $\csc(x)=\frac{\sqrt{13}}{3}$.

If now we substitute these values in the above expression, answer turns out to be $26$. I don't know why this is happening, but please help me find the error in this approach.

Thank you!

Answer 1

Hint:

$2\sec x-3\csc x=0$ may not give us the minimum value as $\sec x\csc x$ is not constant

Use $$4\sec^2x+9\csc^2x=4(1+\tan^2x)+9(1+\cot^2x)=13+(2\tan x-3\cot x)^2+2\cdot2\cdot3$$

Answer 2

substitute $$cos^2(x) = t$$

Having $$sec(x) = \frac{1}{cos(x)}$$ $$cosec(x) = \frac{1}{sin(x)}$$ $$ \frac{4}{cos^2(x)} + \frac {9}{sin^2(x)} -> min, 0<x<\pi$$ $$ \frac{4sin^2(x)}{sin^2(x)cos^2(x)} + \frac{9cos^2(x)}{sin^2(x)cos^2(x)} -> min$$ substitute $ t = cos^2(x), sin^2(x) = 1-t $ $$ \frac{4+5t}{(1-t)t} -> min $$ finding minimum by differentiation $$ \frac{d}{dt} \frac{4+5t}{(1-t)t} = 0$$ $$ \frac{5t^2+8t-4}{t^2(1-t)^2} =0 $$ hence $$ 5t^2+8t-4 = 0, t \ne 0, t \ne 1 $$ solving we get $ t= \frac{2}{5},-2 $ Skipping verification that this value is minimum, we get $$ cos^2(x) = \frac{2}{5} $$ $$ x = arccos (\sqrt(\frac{2}{5}))$$ and minimum value $$ \frac{4+5t}{(1-t)t} = \frac{6}{\frac{3}{5}*\frac{2}{5}} = 25$$

Answer 3

Let the minimum value be $a$. Then:

$$4 \sec^2 x + 9 \csc^2 x = a \implies 4 \sin^2 x + 9 \cos^2 x = a\cos^2x\sin^2x$$ $$\implies 4 - 4 \cos^2 x + 9 \cos^2 x = a \cos^2 x (1 - \cos^2 x)$$ $$\implies -a \cos^4 x + (a - 5) \cos^2 x - 4 = 0$$

and since $\Delta = 0$, $(a - 5)^2 - 4(-a)(-4) = 0$ or $a^2 - 26a + 25 = 0 \implies a =1, 25$. But since $4 \sec^2 x + 9 \csc^2 x ≥ 9 \csc^2 x = \frac{9}{\sin^2 x} ≥ \frac{9}{1}$, the minimum value must be $25$.

To check if the minimum value is attained in the first quadrant, substitute $a = 25$ into the quadratic equation above, which must result in a perfect square as $\Delta = 0$.