જો cos ^{-1} x-cos ^{-1} frac{y}{3}=alpha,જ્ય | vedclass

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(C) આપેલ સમીકરણ $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$ છે. નિત્યસમ $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} (uv + \sqrt{1-u^2}\sqrt{1-v^2})

Vedclass · Accepted Answer

$9\sin ^2 \alpha$

Vedclass · Answer

(C) આપેલ સમીકરણ $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$ છે. નિત્યસમ $\cos ^{-1} u - \cos ^{-1} v = \cos ^{-1} (uv + \sqrt{1-u^2}\sqrt{1-v^2})$ નો ઉપયોગ કરતા,આપણને મળે $\cos ^{-1} (\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}}) = \alpha$. બંને બાજુ કોસાઇન લેતા: $\frac{xy}{3} + \sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}} = \cos \alpha$. પદ ગોઠવતા: $\sqrt{1-x^2}\sqrt{1-\frac{y^2}{9}} = \cos \alpha - \frac{xy}{3}$. બંને બાજુ વર્ગ કરતા: $(1-x^2)(1-\frac{y^2}{9}) = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha + \frac{x^2 y^2}{9}$. $1 - \frac{y^2}{9} - x^2 + \frac{x^2 y^2}{9} = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha + \frac{x^2 y^2}{9}$. $1 - x^2 - \frac{y^2}{9} = \cos^2 \alpha - \frac{2xy}{3} \cos \alpha$. $1 - \cos^2 \alpha = x^2 - \frac{2xy}{3} \cos \alpha + \frac{y^2}{9}$. $\sin^2 \alpha = x^2 - \frac{2xy}{3} \cos \alpha + \frac{y^2}{9}$. $9$ વડે ગુણતા: $9 \sin^2 \alpha = 9x^2 - 6xy \cos \alpha + y^2$.

જો $\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha$,જ્યાં $-1 \leq x \leq 1$,$-3 \leq y \leq 3$,અને $x \leq \frac{y}{3}$ હોય,તો તમામ $x, y$ માટે $9 x^2-6 x y \cos \alpha+y^2$ ની કિંમત શોધો.

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