Question

1+cos2xsinx11+\dfrac{\cos^2 x}{\sin x-1}

Solution

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We are given:

1+cos2xsinx11+\dfrac{\cos^2 x}{\sin x-1}

Use the Pythagorean identity: sin2x+cos2x=1\sin^2 x+\cos^2 x=1

=1+1sin2xsinx1=1+\dfrac{1-\sin^2 x}{\sin x-1}

Factor 1sin2x1-\sin^2 x as difference of two squares:

=1+(1sinx)(1+sinx)sinx1=1+\dfrac{(1-\sin x)(1+\sin x)}{\sin x-1}

Rewrite sinx1\sin x-1 as (1sinx)-(1-\sin x):

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

Cancel 1sinx1-\sin x:

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

=1+1+sinx1=1+\dfrac{1+\sin x}{-1}

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

=11sinx=1-1-\sin x

=sinx(1)=\color{#c34632}-\sin x\color{white}\tag{1}

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