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Question

(B) આપેલ સમીકરણ $	an^{-1}(1 - \alpha) + 	an^{-1}(1 - \beta) = \frac{\pi}{4}$ છે.સૂત્ર $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\frac{x+y}{1-xy}ig

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) આપેલ સમીકરણ $	an^{-1}(1 - \alpha) + 	an^{-1}(1 - \beta) = \frac{\pi}{4}$ છે.સૂત્ર $	an^{-1}x + 	an^{-1}y = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$ નો ઉપયોગ કરતા,આપણને મળે છે $\frac{(1-\alpha)+(1-\beta)}{1-(1-\alpha)(1-\beta)} = 	an\left(\frac{\pi}{4}ight) = 1$.આનું સાદું રૂપ આપતા $2 - (\alpha + \beta) = 1 - (1 - \alpha - \beta + \alpha\beta)$ મળે છે.$2 - \alpha - \beta = \alpha + \beta - \alpha\beta \Rightarrow 2 = 2(\alpha + \beta) - \alpha\beta$.સમીકરણમાં $\beta = \frac{1}{3\alpha}$ મૂકતા:$2 = 2(\alpha + \frac{1}{3\alpha}) - \alpha(\frac{1}{3\alpha}) = 2\alpha + \frac{2}{3\alpha} - \frac{1}{3}$.$2 + \frac{1}{3} = 2\alpha + \frac{2}{3\alpha} \Rightarrow \frac{7}{3} = \frac{6\alpha^2 + 2}{3\alpha}$.$7\alpha = 6\alpha^2 + 2 \Rightarrow 6\alpha^2 - 7\alpha + 2 = 0$.દ્વિઘાત સમીકરણના અવયવ પાડતા: $(2\alpha - 1)(3\alpha - 2) = 0$.તેથી,$\alpha = \frac{1}{2}$ અથવા $\alpha = \frac{2}{3}$.જો $\alpha = \frac{1}{2}$ હોય,તો $\beta = \frac{1}{3(1/2)} = \frac{2}{3}$.જો $\alpha = \frac{2}{3}$ હોય,તો $\beta = \frac{1}{3(2/3)} = \frac{1}{2}$.બંને કિસ્સામાં,$\alpha + \beta = \frac{1}{2} + \frac{2}{3} = \frac{7}{6}$.તેથી,$6(\alpha + \beta) = 6 	imes \frac{7}{6} = 7$.

ધારો કે $0 < \alpha < 1$,$\beta = \frac{1}{3\alpha}$,અને $\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$ છે. તો $6(\alpha + \beta)$ ની કિંમત શોધો:

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