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If $\sec x+\tan x=3, x \in\left(0, \frac{\pi}{2}\right)$ then, $\sin x=$
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$\frac{4}{5}$
We have $\sec x+\tan x=3$ ...(1)
We know that $\sec ^{2} x-\tan ^{2} x=1$
$\therefore(\sec x-\tan x)(\sec x+\tan x)=1 \Rightarrow \sec x-\tan x=\frac{1}{3}$ ...(2)
Adding (1) and (2)
$2 \sec x=\frac{10}{3} \Rightarrow \sec x=\frac{5}{3} \Rightarrow \cos x=\frac{3}{5} \Rightarrow \sin x=\sqrt{1-\cos ^{2} x}=\frac{4}{5}$
We know that $\sec ^{2} x-\tan ^{2} x=1$
$\therefore(\sec x-\tan x)(\sec x+\tan x)=1 \Rightarrow \sec x-\tan x=\frac{1}{3}$ ...(2)
Adding (1) and (2)
$2 \sec x=\frac{10}{3} \Rightarrow \sec x=\frac{5}{3} \Rightarrow \cos x=\frac{3}{5} \Rightarrow \sin x=\sqrt{1-\cos ^{2} x}=\frac{4}{5}$
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