If 1+sqrt{1+a}=(1+sqrt{1-a}) cot alpha and 0 | vedclass

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(A) Given the equation: $(1+\sqrt{1+a})=(1+\sqrt{1-a}) \cot \alpha$ Rewrite as: $(1+\sqrt{1+a}) \sin \alpha = (1+\sqrt{1-a}) \cos \alpha$ Rearranging

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$a$

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(A) Given the equation: $(1+\sqrt{1+a})=(1+\sqrt{1-a}) \cot \alpha$ Rewrite as: $(1+\sqrt{1+a}) \sin \alpha = (1+\sqrt{1-a}) \cos \alpha$ Rearranging terms: $(\sin \alpha - \cos \alpha) = \sqrt{1-a} \cos \alpha - \sqrt{1+a} \sin \alpha$ Squaring both sides: $(\sin \alpha - \cos \alpha)^2 = (\sqrt{1-a} \cos \alpha - \sqrt{1+a} \sin \alpha)^2$ $1 - \sin 2 \alpha = (1-a) \cos^2 \alpha + (1+a) \sin^2 \alpha - 2\sqrt{1-a^2} \sin \alpha \cos \alpha$ $1 - \sin 2 \alpha = (\cos^2 \alpha + \sin^2 \alpha) + a(\sin^2 \alpha - \cos^2 \alpha) - \sqrt{1-a^2} \sin 2 \alpha$ $1 - \sin 2 \alpha = 1 - a \cos 2 \alpha - \sqrt{1-a^2} \sin 2 \alpha$ $a \cos 2 \alpha - \sin 2 \alpha = -\sqrt{1-a^2} \sin 2 \alpha$ Squaring again: $a^2 \cos^2 2 \alpha + \sin^2 2 \alpha - 2a \sin 2 \alpha \cos 2 \alpha = (1-a^2) \sin^2 2 \alpha$ $a^2 \cos^2 2 \alpha + \sin^2 2 \alpha - a \sin 4 \alpha = \sin^2 2 \alpha - a^2 \sin^2 2 \alpha$ $a^2 \cos^2 2 \alpha + a^2 \sin^2 2 \alpha = a \sin 4 \alpha$ $a^2(1) = a \sin 4 \alpha$ Therefore,$\sin 4 \alpha = a$.

If $1+\sqrt{1+a}=(1+\sqrt{1-a}) \cot \alpha$ and $0 < a < 1$,then $\sin 4 \alpha=$

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