If $\cot \theta=\frac{15}{8}, \theta$ is an acute angle, then find the value of $\frac{(1-\cos \theta)(2+2 \cos \theta)}{(2-2 \sin \theta)(1+\sin \theta)}$.
Answer & explanation
Correct answer: option 2
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
= \(\frac{B²}{P²}\)
= \(\frac{8²}{15²}\)
= \(\frac{64}{225}\)