Question

use this method (Pythagoras theorem) to solve @Takenname... ​

Answer 1

Solution is in attachment

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Answer 2

$\huge\text{(b) \dfrac{225}{64} }$

$\large\underline{\large\underline{\text{Explanation}}}$

We know that,

$\cdots\longrightarrow\sin\theta=\dfrac{A}{H},\cos\theta=\dfrac{P}{H}.$

Since,

$\cdots\longrightarrow\cot\theta=\dfrac{\cos\theta}{\sin\theta}$

let us suppose,

$\cdots\longrightarrow A=8k,P=15k.$

By the Pythagorean theorem, the hypotenuse is,

$\cdots\longrightarrow \underline{A^{2}+P^{2}=H^{2}.}$

$\cdots\longrightarrow(8k)^{2}+(15k)^{2}=(17k)^{2}$

$\cdots\longrightarrow \underline{H={17k.}}$

Then,

$\cdots\longrightarrow\sin\theta=\dfrac{8}{17},\cos\theta=\dfrac{15}{17}.$

Given that,

$\dfrac{(2+2\sin\theta)(2-\sin\theta)}{(1+\cos\theta)(2-2\cos\theta)}$

$=\dfrac{2(1+\sin\theta)(1-\sin\theta)}{2(1+\cos\theta)(1-\cos\theta)}$

$=\dfrac{1-\sin^{2}\theta}{1-\cos^{2}\theta}$

$=\dfrac{1-\left(\frac{8}{17}\right)^{2}}{1-\left(\frac{15}{17}\right)^{2}}$

$=\dfrac{\frac{289-64}{289}}{\frac{289-225}{289}}$

$=\dfrac{289-64}{289-225}$

$=\dfrac{225}{64}.$

So,

$\cdots\longrightarrow\boxed{\dfrac{(2+2\sin\theta)(1-1\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\dfrac{225}{64}.}$

Choice (b) is the correct answer.

I hope you are satisfied.