જો cos^{-1}left(frac{x}{a}right) + cos^{-1}le | vedclass

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(A) આપણને આપેલ છે કે $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$. નિત્યસમ $\cos^{-1} A + \cos^{-1} B = \cos^{-1}

Vedclass · Accepted Answer

$\sin^2 \alpha$

Vedclass · Answer

(A) આપણને આપેલ છે કે $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$. નિત્યસમ $\cos^{-1} A + \cos^{-1} B = \cos^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$ નો ઉપયોગ કરતા: $\cos^{-1} \left[ \frac{x}{a} \cdot \frac{y}{b} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} \right] = \alpha$. બંને બાજુ $\cos$ લેતા: $\frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}} = \cos \alpha$. પદોને ગોઠવતા: $\frac{xy}{ab} - \cos \alpha = \sqrt{1 - \frac{x^2}{a^2}} \sqrt{1 - \frac{y^2}{b^2}}$. બંને બાજુ વર્ગ કરતા: $\left( \frac{xy}{ab} - \cos \alpha \right)^2 = \left( 1 - \frac{x^2}{a^2} \right) \left( 1 - \frac{y^2}{b^2} \right)$. $\frac{x^2 y^2}{a^2 b^2} - \frac{2xy}{ab} \cos \alpha + \cos^2 \alpha = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2 y^2}{a^2 b^2}$. બંને બાજુથી $\frac{x^2 y^2}{a^2 b^2}$ ને દૂર કરતા: $-\frac{2xy}{ab} \cos \alpha + \cos^2 \alpha = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2}$. જરૂરી પદ મેળવવા માટે ગોઠવતા: $\frac{x^2}{a^2} - \frac{2xy}{ab} \cos \alpha + \frac{y^2}{b^2} = 1 - \cos^2 \alpha = \sin^2 \alpha$.

જો $\cos^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{y}{b}\right) = \alpha$ હોય,તો $\frac{x^2}{a^2} - \frac{2xy}{ab}\cos \alpha + \frac{y^2}{b^2} = $

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