If cot θ = 15/8, θ is an acute angle, then find the value of ((1-cosθ)(2+2cosθ))/((2-2sinθ)(1+sinθ))

If cot θ = 15/8, θ is an acute angle, then find the value…

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

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Correct answer: B

Explanation

The correct answer is option 2 i.e. $\frac{64}{225}$

cot $\theta$ = 15/8 = B/P

Hypotenuse = $\sqrt{15^2 +8^2}$ = 17

Sin $\theta$ = P/H = 8/17

Cos $\theta$ = B/H = 15/17

= [(1 - 15/17)(2 + 2 × 15/17)]/[(2 - 2 × 8/17)(1 + 8/17)]

= 64/225