જો (1+sqrt{1+x}) tan x=1+sqrt{1-x} હોય,તો sin | vedclass

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(A) આપેલ સમીકરણ: $(1+\sqrt{1+x}) 	an x = 1+\sqrt{1-x}$ $	an x = \frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$ ધારો કે $x = \sin 	heta$. $	an x = \frac{1+\sqr

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$x$

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(A) આપેલ સમીકરણ: $(1+\sqrt{1+x}) 	an x = 1+\sqrt{1-x}$ $	an x = \frac{1+\sqrt{1-x}}{1+\sqrt{1+x}}$ ધારો કે $x = \sin 	heta$. $	an x = \frac{1+\sqrt{1-\sin 	heta}}{1+\sqrt{1+\sin 	heta}} = \frac{1+\sqrt{(\cos \frac{	heta}{2} - \sin \frac{	heta}{2})^2}}{1+\sqrt{(\cos \frac{	heta}{2} + \sin \frac{	heta}{2})^2}}$ $= \frac{1+\cos \frac{	heta}{2} - \sin \frac{	heta}{2}}{1+\cos \frac{	heta}{2} + \sin \frac{	heta}{2}}$ $= \frac{2\cos^2 \frac{	heta}{4} - 2\sin \frac{	heta}{4}\cos \frac{	heta}{4}}{2\cos^2 \frac{	heta}{4} + 2\sin \frac{	heta}{4}\cos \frac{	heta}{4}}$ $= \frac{2\cos \frac{	heta}{4}(\cos \frac{	heta}{4} - \sin \frac{	heta}{4})}{2\cos \frac{	heta}{4}(\cos \frac{	heta}{4} + \sin \frac{	heta}{4})} = \frac{1 - 	an \frac{	heta}{4}}{1 + 	an \frac{	heta}{4}} = 	an(\frac{\pi}{4} - \frac{	heta}{4})$ તેથી,$x = \frac{\pi}{4} - \frac{	heta}{4} \Rightarrow 4x = \pi - 	heta$. $\sin 4x = \sin(\pi - 	heta) = \sin 	heta = x$.

જો $(1+\sqrt{1+x}) \tan x=1+\sqrt{1-x}$ હોય,તો $\sin 4x$ ની કિંમત શું થાય?

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