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

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(A) આપેલ છે $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. નિત્યસમ $\cos ^{-1} a - \cos ^{-1} b = \cos ^{-1} (ab + \sqrt{1-a^2} \sqrt{1-b^2})$ નો ઉ

Vedclass · Accepted Answer

$9 \sin ^2 \alpha$

Vedclass · Answer

(A) આપેલ છે $\cos ^{-1} x - \cos ^{-1} \frac{y}{3} = \alpha$. નિત્યસમ $\cos ^{-1} a - \cos ^{-1} b = \cos ^{-1} (ab + \sqrt{1-a^2} \sqrt{1-b^2})$ નો ઉપયોગ કરતા: $\cos ^{-1} \left( \frac{xy}{3} + \sqrt{1-x^2} \sqrt{1-\frac{y^2}{9}} \right) = \alpha$ $\frac{xy}{3} + \frac{\sqrt{1-x^2} \sqrt{9-y^2}}{3} = \cos \alpha$ $xy + \sqrt{1-x^2} \sqrt{9-y^2} = 3 \cos \alpha$ $xy - 3 \cos \alpha = -\sqrt{1-x^2} \sqrt{9-y^2}$ બંને બાજુ વર્ગ કરતા: $(xy - 3 \cos \alpha)^2 = (1-x^2)(9-y^2)$ $x^2y^2 - 6xy \cos \alpha + 9 \cos ^2 \alpha = 9 - y^2 - 9x^2 + x^2y^2$ $9x^2 - 6xy \cos \alpha + y^2 = 9 - 9 \cos ^2 \alpha$ $9x^2 - 6xy \cos \alpha + y^2 = 9(1 - \cos ^2 \alpha) = 9 \sin ^2 \alpha$.

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 $9x^2 - 6xy \cos \alpha + y^2$ ની કિંમત શું થાય?

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