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stellarik [79]
4 years ago
13

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

Mathematics
1 answer:
BARSIC [14]4 years ago
6 0

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

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)

On the left side, we can simplify a bit:

\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}

\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}

Then

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)

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

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s

Combine the fractions:

\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{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

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s

as desired.

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