If sec x + tan x = 3, x in (0, frac{pi}{2}),t | vedclass

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(B) Given $\sec x + 	an x = 3$  $(1)$We know that $\sec^2 x - 	an^2 x = 1$Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(\sec x - 	an x)(

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$\frac{4}{5}$

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(B) Given $\sec x + 	an x = 3$  $(1)$We know that $\sec^2 x - 	an^2 x = 1$Using the identity $a^2 - b^2 = (a - b)(a + b)$,we get $(\sec x - 	an x)(\sec x + 	an x) = 1$Substituting $(1)$,we have $\sec x - 	an x = \frac{1}{3}$  $(2)$Adding $(1)$ and $(2)$,we get $2 \sec x = 3 + \frac{1}{3} = \frac{10}{3}$Thus,$\sec x = \frac{5}{3}$,which implies $\cos x = \frac{3}{5}$Since $x \in (0, \frac{\pi}{2})$,$\sin x$ is positive.$\sin x = \sqrt{1 - \cos^2 x} = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$

If $\sec x + \tan x = 3, x \in (0, \frac{\pi}{2})$,then $\sin x =$

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