If cos ^{-1} 2x + cos ^{-1} 3x = frac{pi}{3} | vedclass

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(D) Given: $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$ Taking $\cos$ on both sides: $\cos(\cos ^{-1} 2x + \cos ^{-1} 3x) = \cos(\frac{\pi}{3})$ Us

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$10$

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(D) Given: $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$ Taking $\cos$ on both sides: $\cos(\cos ^{-1} 2x + \cos ^{-1} 3x) = \cos(\frac{\pi}{3})$ Using the formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $(2x)(3x) - \sqrt{1-(2x)^2} \sqrt{1-(3x)^2} = \frac{1}{2}$ $6x^2 - \frac{1}{2} = \sqrt{1-4x^2} \sqrt{1-9x^2}$ Squaring both sides: $(6x^2 - \frac{1}{2})^2 = (1-4x^2)(1-9x^2)$ $36x^4 - 6x^2 + \frac{1}{4} = 1 - 13x^2 + 36x^4$ $7x^2 = \frac{3}{4}$ $x^2 = \frac{3}{28}$ Then $4x^2 = 4 	imes \frac{3}{28} = \frac{3}{7} = \frac{a}{b}$ Thus,$a = 3$ and $b = 7$. $a + b = 3 + 7 = 10$.

If $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$ and $4x^2 = \frac{a}{b}$,then $a + b =$

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