Question

Let (3 pi)/(4)<theta<pi and sqrt(2cot theta+(1)/(sin^(2)theta))=K-cot theta then K equals (A) -1 (B) 0 (C) 1/2 (D) 1
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Answer 1

Step-by-step explanation:

We have,

$= > \sqrt{2 \cot( \theta) + \frac{1}{ \sin^{2} (\theta) } } = k - \cot(\theta)$

$= > \sqrt{2 \frac{ \cos( \theta ) }{ \sin( \theta) } + \frac{1}{ \sin^{2} (\theta) } } = k - \cot(\theta)$

$= > \sqrt{ \frac{2 \sin( \theta) \cos( \theta) + 1}{ \sin^{2} ( \theta) } } = k - \cot(\theta)$

$= > \sqrt{ \frac{( \sin( \theta) + \cos( \theta) )^{2} }{ \sin^{2} ( \theta) } } = k - \cot(\theta)$

$= > \frac{ - ( \cos( \theta) + \sin( \theta) )}{ \sin( \theta) } = k - \cot(\theta)$

$= > k = - 1$