If sec x + tan x = 2,where 0

Question

(B) Given $\sec x + 	an x = 2$. We know that $\sec^2 x - 	an^2 x = 1$,which implies $(\sec x - 	an x)(\sec x + 	an x) = 1$. Substituting the given

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$\frac{1}{\sqrt{2(10+3 \sqrt{10})}}$

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(B) Given $\sec x + 	an x = 2$. We know that $\sec^2 x - 	an^2 x = 1$,which implies $(\sec x - 	an x)(\sec x + 	an x) = 1$. Substituting the given value,we get $\sec x - 	an x = \frac{1}{2}$. Adding the two equations: $2 \sec x = 2 + \frac{1}{2} = \frac{5}{2}$,so $\sec x = \frac{5}{4}$. Thus,$\cos x = \frac{4}{5}$. Using the half-angle formula $\sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} = \frac{1 - 4/5}{2} = \frac{1}{10}$. Since $0 Then $\cos \frac{x}{2} = \sqrt{1 - \sin^2 \frac{x}{2}} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$. Using $\sin^2 \frac{x}{4} = \frac{1 - \cos \frac{x}{2}}{2} = \frac{1 - 3/\sqrt{10}}{2} = \frac{\sqrt{10} - 3}{2 \sqrt{10}}$. To rationalize,multiply numerator and denominator by $\sqrt{10} + 3$: $\sin^2 \frac{x}{4} = \frac{(\sqrt{10} - 3)(\sqrt{10} + 3)}{2 \sqrt{10}(\sqrt{10} + 3)} = \frac{10 - 9}{2(10 + 3 \sqrt{10})} = \frac{1}{2(10 + 3 \sqrt{10})}$. Therefore,$\sin \frac{x}{4} = \frac{1}{\sqrt{2(10 + 3 \sqrt{10})}}$.

If $\sec x + \tan x = 2$,where $0 < x < \frac{\pi}{2}$,then find the value of $\sin \frac{x}{4}$.

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