For the functions f(theta) = alpha tan^2 thet | vedclass

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(NONE) The minimum value of $f(	heta) = \alpha 	an^2 	heta + \beta \cot^2 	heta$ is found using the $AM$-$GM$ inequality: $f(	heta) \ge 2\sqrt{\a

Vedclass · Accepted Answer

$1023$

Vedclass · Answer

(NONE) The minimum value of $f(	heta) = \alpha 	an^2 	heta + \beta \cot^2 	heta$ is found using the $AM$-$GM$ inequality: $f(	heta) \ge 2\sqrt{\alpha 	an^2 	heta \cdot \beta \cot^2 	heta} = 2\sqrt{\alpha\beta}$.
Since $\alpha > \beta > 0$, the maximum value of $g(	heta) = \alpha \sin^2 	heta + \beta \cos^2 	heta$ is $\alpha$.
Given $\min f(	heta) = \max g(	heta)$, we have $2\sqrt{\alpha\beta} = \alpha$.
Squaring both sides, $4\alpha\beta = \alpha^2$, which implies $\alpha = 4\beta$ (since $\alpha 
eq 0$).
The first term $a = \frac{\alpha}{2\beta} = \frac{4\beta}{2\beta} = 2$.
The common ratio $r = \frac{2\beta}{\alpha} = \frac{2\beta}{4\beta} = \frac{1}{2}$.
The sum of the first $10$ terms of the $G$.$P$. is $S_{10} = a \frac{1-r^{10}}{1-r} = 2 \frac{1-(1/2)^{10}}{1-1/2} = 4(1 - \frac{1}{1024}) = 4 \cdot \frac{1023}{1024} = \frac{1023}{256}$.
Thus, $m = 1023$ and $n = 256$. Since $\gcd(1023, 256) = 1$, $m+n = 1023 + 256 = 1279$.

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