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Question Number 162728 by mkam last updated on 31/Dec/21

Commented by mkam last updated on 31/Dec/21

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Commented by tounghoungko last updated on 01/Jan/22
$f(x)=(a^2 /u)+(b^2 /(1−u)) ; u=sin^2 x ; 0<u<1 f ′(x)=[−(a^2 /u^2 )+(b^2 /((1−u)^2 )) ].sin 2x =0 ⇔ (bu)^2 −(a−au)^2 = 0 ⇔ ((b−a)u+a)((b+a)u−a)=0 { ((u=−(a/(b−a))=(a/(a−b)))),((u=(a/(a+b)))) :} f(x)_(min) = (a^2 /(((a/(a+b))))) + (b^2 /((1−(a/(a+b))))) = a(a+b)+b(a+b)= (a+b)^2 =∣a+b∣^2$
$f (x) = \frac{a^{2}}{u} + \frac{b^{2}}{1 - u}; u = \sin^{2} x; 0 < u < 1$ $f^{'} (x) = [- \frac{a^{2}}{u^{2}} + \frac{b^{2}}{{(1 - u)}^{2}}] . \sin 2 x = 0$ $\Leftrightarrow {(b u)}^{2} - {(a - a u)}^{2} = 0$ $\Leftrightarrow ((b - a) u + a) ((b + a) u - a) = 0$ ${\begin{cases} u = - \frac{a}{b - a} = \frac{a}{a - b} \\ u = \frac{a}{a + b} \end{cases}$ $f {(x)}_{m i n} = \frac{a^{2}}{(\frac{a}{a + b})} + \frac{b^{2}}{(1 - \frac{a}{a + b})}$ $= a (a + b) + b (a + b) = {(a + b)}^{2} =∣ a + b ∣^{2}$
	Answered by mnjuly1970 last updated on 31/Dec/21

	$f (x) = a^{2} + b^{2} + a^{2} {c o t}^{2} (x) + b^{2} {t a n}^{2} (x)$ $⩾ a^{2} + b^{2} + 2 \sqrt{a^{2} . b^{2}} = {(∣ a ∣ + ∣ b ∣)}^{2}$ $f_{m i n} = {(∣ a ∣ + ∣ b ∣)}^{2}$
	Answered by safmoh22 last updated on 31/Dec/21

	$665 n [x \frac{66 x}{[22]} b]$
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