જો S = {x in R : sin^{-1}left(frac{x+1}{sqrt{ | vedclass

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(C) ધારો કે $f(x) = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}$.ધારો કે $\alpha

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$4$

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(C) ધારો કે $f(x) = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}$.ધારો કે $\alpha = \sin^{-1}\left(\frac{x+1}{\sqrt{(x+1)^2+1}}\right)$ અને $\beta = \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)$.તેથી $\sin \alpha = \frac{x+1}{\sqrt{(x+1)^2+1}}$ અને $\sin \beta = \frac{x}{\sqrt{x^2+1}}$.કારણ કે $\cos \alpha = \frac{1}{\sqrt{(x+1)^2+1}}$ અને $\cos \beta = \frac{1}{\sqrt{x^2+1}}$,આપણને મળે $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \frac{1}{\sqrt{(x+1)^2+1}\sqrt{x^2+1}}$.આપેલ છે કે $\alpha - \beta = \frac{\pi}{4}$,તેથી $\sin(\alpha - \beta) = \frac{1}{\sqrt{2}}$.આમ,$\frac{1}{\sqrt{(x+1)^2+1}\sqrt{x^2+1}} = \frac{1}{\sqrt{2}} \implies ((x+1)^2+1)(x^2+1) = 2$.$(x^2+2x+2)(x^2+1) = 2 \implies x^4 + 2x^3 + 3x^2 + 2x + 2 = 2 \implies x(x+1)(x^2+x+2) = 0$.વાસ્તવિક ઉકેલો $x=0$ અને $x=-1$ મળે છે.$S = \{-1, 0\}$.$x=0$ માટે,$\sin(5\pi/2) - \cos(5\pi) = 1 - (-1) = 2$.$x=-1$ માટે,$\sin(5\pi/2) - \cos(5\pi) = 1 - (-1) = 2$.સરવાળો $= 2 + 2 = 4$.

જો $S = \{x \in R : \sin^{-1}\left(\frac{x+1}{\sqrt{x^2+2x+2}}\right) - \sin^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right) = \frac{\pi}{4}\}$,હોય તો $\sum_{x \in S} \left(\sin\left((x^2+x+5)\frac{\pi}{2}\right) - \cos((x^2+x+5)\pi)\right)$ ની કિંમત $........$ 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.

જો $\sin^{-1} \frac{3}{5} + \cos^{-1} \frac{12}{13} = \sin^{-1} C$ હોય,તો $C =$

$\lim _{x \rightarrow 0^{+}} \frac{x \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)}{\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) \tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right)}$ ની કિંમત શોધો.

જો $\cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1-p} + \cos^{-1} \sqrt{1-q} = \frac{3\pi}{4}$ હોય,તો $q$ ની કિંમત શોધો.

$ \cos \left[2 \sin ^{-1} \frac{3}{4} + \cos ^{-1} \frac{3}{4}\right] $

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