જો cos^{-1} left( frac{x}{2} right) + cos^{-1 | vedclass

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(A) આપેલ છે: $\cos^{-1} \left( \frac{x}{2} ight) + \cos^{-1} \left( \frac{y}{3} ight) = 	heta$ નિત્યસમ $\cos^{-1} A + \cos^{-1} B = \cos^{-1} \le

Vedclass · Accepted Answer

$36 \sin^{2} 	heta$

Vedclass · Answer

(A) આપેલ છે: $\cos^{-1} \left( \frac{x}{2} ight) + \cos^{-1} \left( \frac{y}{3} ight) = 	heta$ નિત્યસમ $\cos^{-1} A + \cos^{-1} B = \cos^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} ight)$ નો ઉપયોગ કરતા: $\cos^{-1} \left\{ \left( \frac{x}{2} ight) \left( \frac{y}{3} ight) - \sqrt{1 - \frac{x^{2}}{4}} \sqrt{1 - \frac{y^{2}}{9}} ight\} = 	heta$ $\Rightarrow \frac{xy}{6} - \sqrt{\frac{4-x^2}{4}} \sqrt{\frac{9-y^2}{9}} = \cos 	heta$ $\Rightarrow \frac{xy}{6} - \frac{\sqrt{4-x^2} \sqrt{9-y^2}}{6} = \cos 	heta$ $\Rightarrow xy - \sqrt{4-x^2} \sqrt{9-y^2} = 6 \cos 	heta$ $\Rightarrow xy - 6 \cos 	heta = \sqrt{4-x^2} \sqrt{9-y^2}$ બંને બાજુ વર્ગ કરતા: $(xy - 6 \cos 	heta)^2 = (4-x^2)(9-y^2)$ $x^2 y^2 - 12xy \cos 	heta + 36 \cos^2 	heta = 36 - 4y^2 - 9x^2 + x^2 y^2$ બંને બાજુથી $x^2 y^2$ દૂર કરતા: $-12xy \cos 	heta + 36 \cos^2 	heta = 36 - 9x^2 - 4y^2$ પદોને ગોઠવતા: $9x^2 + 4y^2 - 12xy \cos 	heta = 36 - 36 \cos^2 	heta$ $9x^2 + 4y^2 - 12xy \cos 	heta = 36(1 - \cos^2 	heta)$ $9x^2 + 4y^2 - 12xy \cos 	heta = 36 \sin^2 	heta$

જો $\cos^{-1} \left( \frac{x}{2} \right) + \cos^{-1} \left( \frac{y}{3} \right) = \theta$ હોય,તો $9 x^{2} - 12 x y \cos \theta + 4 y^{2} =$ શું થાય?

Similar Questions

જો $0 \leqslant \cos ^{-1} x \leqslant \pi$ અને $\frac{-\pi}{2} \leqslant \sin ^{-1} x \leqslant \frac{\pi}{2}$ હોય,તો $x=\frac{1}{5}$ માટે $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ ની કિંમત શોધો.

કિંમત શોધો: $\sec^{-1} x + \operatorname{cosec}^{-1} x + \cos^{-1}(x^{-1}) + \sin^{-1}(x^{-1})$ (જ્યાં $|x| > 1, x \in R$)

જો $0 < x < 1$ હોય,તો $\cot ^{-1}\left( \frac{2x^2 - 1}{2x\sqrt{1 - x^2}} \right)$ ની કિંમત શોધો.

જો $\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$ હોય,તો $x=$ . . . . . . .

$\cos (\tan ^{-1}(\tan 2))$ ની કિંમત શોધો.

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