psinθcon^2 θ=a pcosθsin^2 θ=b p≠0 θ∈(0 ,(π/2)) prove p=(((a^2 +b^2 )^(3/2) )/(ab))


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Question Number 202295 by liuxinnan last updated on 24/Dec/23

$p s i n θ {c o n}^{2} θ = a$ $p c o s θ {s i n}^{2} θ = b$ $p \neq 0 θ \in (0, \frac{π}{2})$ $p r o v e p = \frac{{(a^{2} + b^{2})}^{\frac{3}{2}}}{a b}$
	Answered by 1990mbodji last updated on 24/Dec/23

	$\underset{―}{E x e r c i c e} :$ $p s i n θ {c o s}^{2} θ = a e t p c o s θ {s i n}^{2} θ = b o \overset{`}{u} θ \in] 0; \frac{π}{2} [.$ $M o n t r o n s q u e p = \frac{{(a^{2} + b^{2})}^{\frac{3}{2}}}{a b} .$ $a b = p^{2} {s i n}^{3} θ {c o s}^{3} θ e t a^{2} + b^{2} = p^{2} {s i n}^{2} θ {c o s}^{2} θ ({c o s}^{2} θ + {s i n}^{2} θ)$ $\Rightarrow a b p = {(p s i n θ c o s θ)}^{3} e t a^{2} + b^{2} = {(p s i n θ c o s θ)}^{2}$ $\Rightarrow p s i n θ c o s θ = {(a b p)}^{\frac{1}{3}} e t p s i n θ c o s θ = {(a^{2} + b^{2})}^{\frac{1}{2}}$ $\Rightarrow a b p = {(a^{2} + b^{2})}^{\frac{3}{2}} \Rightarrow p = \frac{{(a^{2} + b^{2})}^{\frac{3}{2}}}{a b} .$
	Commented by MathematicalUser2357 last updated on 26/Dec/23

	$no french$
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