If sin theta + cos theta = frac{17}{13}, 0

Question

(D) Given: $\sin 	heta + \cos 	heta = \frac{17}{13}$ ..... $(1)$Let $\sin 	heta - \cos 	heta = x$ ..... $(2)$Squaring both equations:$(\sin 	heta

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$\frac{7}{13}$

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(D) Given: $\sin 	heta + \cos 	heta = \frac{17}{13}$ ..... $(1)$Let $\sin 	heta - \cos 	heta = x$ ..... $(2)$Squaring both equations:$(\sin 	heta + \cos 	heta)^2 = (\frac{17}{13})^2 \Rightarrow \sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta = \frac{289}{169}$Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $1 + 2 \sin 	heta \cos 	heta = \frac{289}{169} \Rightarrow 2 \sin 	heta \cos 	heta = \frac{289}{169} - 1 = \frac{120}{169}$ ..... $(3)$Now,$x^2 = (\sin 	heta - \cos 	heta)^2 = \sin^2 	heta + \cos^2 	heta - 2 \sin 	heta \cos 	heta$$x^2 = 1 - 2 \sin 	heta \cos 	heta$Substituting the value from $(3)$:$x^2 = 1 - \frac{120}{169} = \frac{169 - 120}{169} = \frac{49}{169}$$x = \pm \sqrt{\frac{49}{169}} = \pm \frac{7}{13}$Since $0  1$,the value is $\frac{7}{13}$.

If $\sin \theta + \cos \theta = \frac{17}{13}, 0 < \theta < 90^{\circ},$ then the value of $\sin \theta - \cos \theta$ is

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