[tex] \large \mathcal\colorbox{red}{SOLUTION:} [/tex]
[tex] \begin{array}{l} \bold{Given:}\:\sf\sin x - \cos x = \dfrac{1}{3} \\ \\ \bold{Required:}\textsf{ Value of }\sf \dfrac{\sin x}{\sec x} \\ \\ \sf \dfrac{\sin x}{\sec x} = \sin x \cos x \\ \\ \sf \sin x - \cos x = \dfrac{1}{3} \\ \\ \textsf{Square both sides.} \\ \\ \Longrightarrow \sf (\sin x - \cos x)^2 = \left(\dfrac{1}{3}\right)^2 \\ \\ \Longrightarrow \sf \sin^2 x - 2\sin x \cos x + \cos^2 x = \dfrac{1}{9} \\ \\ \textsf{By Pythagorean Identity }\sf{\sin^2 \theta + \cos^2 \theta = 1,} \\ \\ \Longrightarrow \sf 1 - 2\sin x \cos x = \dfrac{1}{9} \\ \\ \Longrightarrow \sf 2\sin x \cos x = 1 - \dfrac{1}{9} \\ \\ \Longrightarrow \sf 2\sin x \cos x = \dfrac{8}{9} \\ \\ \Longrightarrow \sf \dfrac{1}{\cancel{2}}(\cancel{2}\sin x \cos x) = \dfrac{1}{2}\!\left(\dfrac{8}{9}\right) \\ \\ \Longrightarrow \sf \sin x \cos x = \boxed{\sf \dfrac{\sin x}{\sec x} = \frac{4}{9}}\quad\textit{Answer} \end{array} [/tex]
[tex] \mathfrak\colorbox{blue}{\#CarryOnLearning} [/tex]
