Question

# $1+\dfrac{\cos^2 x}{\sin x-1}$

Solution

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Step 1
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We are given:

$1+\dfrac{\cos^2 x}{\sin x-1}$

Use the Pythagorean identity: $\sin^2 x+\cos^2 x=1$

$=1+\dfrac{1-\sin^2 x}{\sin x-1}$

Factor $1-\sin^2 x$ as difference of two squares:

$=1+\dfrac{(1-\sin x)(1+\sin x)}{\sin x-1}$

Rewrite $\sin x-1$ as $-(1-\sin x)$:

$=1+\dfrac{(1-\sin x)(1+\sin x)}{-(1-\sin x)}$

Cancel $1-\sin x$:

$=1+\dfrac{\cancel{(1-\sin x)}(1+\sin x)}{-\cancel{(1-\sin x)}}$

$=1+\dfrac{1+\sin x}{-1}$

$=1-(1+\sin x)$

$=1-1-\sin x$

$=\color{#c34632}-\sin x\color{white}\tag{1}$

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