If $\cot \theta=\frac{15}{8}, \theta$ is an acute angle, then find the value of $\frac{(1-\cos \theta)(2+2 \cos \theta)}{(2-2 \sin \theta)(1+\sin \theta)}$.

$\frac{64}{225}$

cotθ = \(\frac{15}{8}\)

{ we know, cotA = \(\frac{B}{P}\) }

By using pythagoras theorem,

P² + B² = H²

8² + 15² = H²

H = 17

Now,

\(\frac{ (1 - cosθ). (2+2cosθ) }{(2-2sinθ). (1 + sinθ ) }\)

= \(\frac{ (1 - cosθ). (1+cosθ) }{(1-sinθ). (1 + sinθ ) }\)

= \(\frac{ (1 - cos²θ) }{(1-sin²θ) }\)

{ we know, sin²θ + cos²θ = 1 }

= \(\frac{ (sin²θ) }{(cos²θ) }\)

= tan²θ

= \(\frac{B²}{P²}\)

= \(\frac{8²}{15²}\)

= \(\frac{64}{225}\)