If cot θ = \(\frac{13}{12}\), θ is an acute angle, then find value of \(\frac{(1 - cos θ) (2 + 2 cosθ)}{(2 - 2sinθ) (1 + sinθ)}\)

\(\frac{144}{169}\)

\(\frac{(1 - cos θ) (2 + 2 cosθ)}{(2 - 2sinθ) (1 + sinθ)}\) = \(\frac{2(1 - cos θ) (1 + cosθ)}{2(1 - sinθ) (1 + sinθ)}\)

= \(\frac{1 - cos^2 θ}{1- sin^2 θ}\) = \(\frac{sin^2 θ }{cos^2 θ}\) = tan² θ

Now, ATQ

⇒ cot θ = \(\frac{13}{12}\) 

⇒ tan θ = \(\frac{12}{13}\)

⇒ tan² θ = \(\frac{144}{169}\)