Question

Evaluate it


$cot \: = \: \frac{15}{8} \\ \\ \frac{(2 \: + \: 2sin)(1 \: - \: sin)}{(1 + cos)(2 - 2cos)}$

Answer 1

HELLO DEAR,

GIVEN:-
cotθ = 15/8

so, $\bold{\qquad \frac{(2 + 2sin\Theta)(1 - sin\Theta)}{(1 + cos\Theta)(2 - 2cos\Theta)}}$

$\bold{\qquad \Rightarrow \frac{2(1 + sin\Theta)(1 - sin\Theta)}{2(1 + cos\Theta)(1 - cos\Theta)}}$

$\bold{\qquad \Rightarrow \frac{(1 - sin^2\Theta)}{(1 - cos^2\Theta)}}$

$\bold{\qquad \Rightarrow \frac{cos^2\Theta}{sin^2\Theta}}$

$\bold{\qquad \Rightarrow cot^2\Theta}$

[as cotθ = 15/8 so, cot²θ = 225/64]

$\bold{\therefore, \Rightarrow cot^2\Theta = 225/64}$

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hi there ♥️

Look at the attachment :)

Hence the value is 225/64.

Thanks !!.