Given tanθ=815\tan\theta = \frac{8}{15}tanθ=158
sinθ=817,cosθ=1517\sin \theta = \frac{8}{17}, \quad \cos \theta = \frac{15}{17}sinθ=178,cosθ=1715 from pythagora's theorem.
To find sinθ−cosθsin2θ−sinθ\frac{\sin \theta - \cos \theta}{\sin^2 \theta - \sin \theta}sin2θ−sinθsinθ−cosθ
Numerator → sinθ−cosθ=817−1517=−717\sin \theta - \cos \theta = \frac{8}{17} - \frac{15}{17} = \frac{-7}{17}sinθ−cosθ=178−1715=17−7
Denominator → sin2θ=(817)2=64289,sin2θ−sinθ=64289−136289=−72289\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}sin2θ=(178)2=28964,sin2θ−sinθ=28964−289136=289−72
−717−72289=7×28917×72=11972\frac{\frac{-7}{17}}{\frac{-72}{289}} = \frac{7 \times 289}{17 \times 72} = \frac{119}{72}289−7217−7=17×727×289=72119