Question

If cot theta= 15/8. Evaluate-
(2 + 2 sin theta)(1- sin theta) / (1 + cos theta)(2- 2 cos theta)

Answer 1

here is your answer

hope its help you

Answer 2

Answer:

$\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}=\frac{225}{64}$

Step-by-step explanation:

By using Pythagoras theorem, (in a right-angled triangle, squaring the longest side is eventually equals to sum of squares of other two sides.)

$\sin \theta=\frac{\text {Opposite}}{\text {Hypotenuse}}$

$\cos \theta=\frac{\text {Adjacent}}{\text {Hypotenuse}}$

$\cot \theta=\frac{\text {Adjacent}}{\text {Opposite}}$

$\cot \theta=\frac{15}{8}$

$\text{Hypotenuse}^{2}=\text { Opposite }^{2}+\text { Adjacent }^{2}$

$H^{2}=O^{2}+A^{2}$

$\Rightarrow 8^{2}+15^{2}$

$\Rightarrow 64+225$

$H^{2} \Rightarrow 289$

$H=\sqrt{289}$

$H=17$

Then, the values of $\sin \theta$ and $\cos \theta$ are,

$\sin \theta=\frac{\text {Opposite}}{\text {Hypotenuse}}$

$\sin \theta=\frac{8}{17}$

$\cos \theta=\frac{\text {Adjacent}}{\text {Hypotenuse}}$

$\cos \theta=\frac{15}{17}$

Then, putting values of \sin \theta and \cos \theta,

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

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

$=\frac{\left(2+\frac{16}{17}\right)\left(\frac{9}{17}\right)}{\left(\frac{32}{17}\right)\left(2-\frac{30}{17}\right)}$

$=\frac{\left(\frac{50}{17}\right)\left(\frac{9}{17}\right)}{\left(\frac{32}{17}\right)\left(\frac{4}{17}\right)}$

$=\frac{\left(\frac{450}{289}\right)}{\left(\frac{128}{289}\right)}$

$=\frac{450}{289} \times \frac{289}{128}$

$=\frac{450}{128}$

$\therefore \frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}=\frac{225}{64}$