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Can you prove it??
it's hard,I tried but couldn't solve this.

Can you prove it?? it's hard,I tried but couldn't solve this.-example-1

1 Answer

1 vote

I'll abbreviate
s=\sin\theta and
c=\cos\theta, so the identity to prove is


(s+c+1)/(s+c-1)-(1+s-c)/(1-s+c)=2\left(1+\frac1s\right)

On the left side, we can simplify a bit:


(s+c+1)/(s+c-1)=(s+c-1+2)/(s+c-1)=1+\frac2{s+c-1}


(1+s-c)/(1-s+c)=-(-2+1-s+c)/(1-s+c)=-1+\frac2{1-s+c}

Then


(s+c+1)/(s+c-1)-(1+s-c)/(1-s+c)=2\left(1+\frac1{s+c-1}-\frac1{1-s+c}\right)

So the establish the original equality, we need to show that


\frac1{s+c-1}-\frac1{1-s+c}=\frac1s

Combine the fractions:


((1-s+c)-(s+c-1))/((s+c-1)(1-s+c))=(2-2s)/(c^2-s^2+2s-1)

We can rewrite the denominator as


c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1

then using the fact that
c^2+s^2=\cos^2\theta+\sin^2\theta=1, we get


1-2s^2+2s-1=2s-2s^2

so that we have


\frac1{s+c-1}-\frac1{1-s+c}=(2-2s)/(2s-2s^2)=\frac1s

as desired.

User XOneca
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