If tan\(\theta\) = \(\frac{8}{15}\), simplify \(\frac{ Sin\theta – Cos\theta}{Sin^2\theta – Sin\theta}\)

Explanation

Given \(\tan \theta = \frac{8}{15}\)

\[\sin \theta = \frac{8}{17}, \quad \cos \theta = \frac{15}{17}\] from pythagora's theorem.

To find  \[\frac{\sin \theta - \cos \theta}{\sin^2 \theta - \sin \theta}\]

Numerator  → \[\sin \theta - \cos \theta = \frac{8}{17} - \frac{15}{17} = \frac{-7}{17}\] 

Denominator → \[\sin^2 \theta = \left(\frac{8}{17}\right)^2 = \frac{64}{289}, \quad \sin^2 \theta - \sin \theta = \frac{64}{289} - \frac{136}{289} = \frac{-72}{289}\] 

\[\frac{\frac{-7}{17}}{\frac{-72}{289}} = \frac{7 \times 289}{17 \times 72} = \frac{119}{72}\]

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