The value of tan left{frac{1}{2} cos ^{-1} fr | vedclass

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Question

(A) Let $	heta = \frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}$.Then $2	heta = \cos ^{-1} \frac{\sqrt{5}}{3}$,which implies $\cos 2	heta = \frac{\sqrt{

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$\frac{3-\sqrt{5}}{2}$

Vedclass · Answer

(A) Let $	heta = \frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}$.Then $2	heta = \cos ^{-1} \frac{\sqrt{5}}{3}$,which implies $\cos 2	heta = \frac{\sqrt{5}}{3}$.We need to find $	an 	heta$.Using the formula $	an 	heta = \sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}}$:$	an 	heta = \sqrt{\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}$.Rationalizing the denominator:$	an 	heta = \sqrt{\frac{(3-\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}} = \sqrt{\frac{(3-\sqrt{5})^2}{9-5}} = \sqrt{\frac{(3-\sqrt{5})^2}{4}} = \frac{3-\sqrt{5}}{2}$.

The value of $\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}$ is

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