How to prove that \[ \frac12\frac1{1+\sin^2 x} + \frac12\frac1{1+\cos^2 x} + \frac12\frac1{1+\sec^2 x}+ \frac12\frac1{1+\csc^2 x} = 1? \] Some genius please help me I have been stuck at this for one whole day.
A sum of terms like $(1 + \sin^2x)^{-1}$
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trigonometry
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1I would start by getting rid of $\sec$ and $\csc$, replacing them with $\frac 1\cos$ and $\frac 1\sin$. Then you must have identities for $1+\sin^2 x$ and $1 + \cos^2 x$. Where does that take you? – 2012-07-24
2 Answers
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Forget about the $2$'s for a while. We have $\frac{1}{1+\sec^2 x}=\frac{1}{1+\frac{1}{\cos^2 x}}=\frac{\cos^2 x}{\cos^2 x+1}.$ Now add $\frac{1}{1+\cos^2 x}$. We get something very simple. Continue.
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Observe that for any $\alpha> 0$ we have $ \frac{1}{1+\alpha}+\frac{1}{1+1/\alpha}=\frac{1+1/\alpha+1+\alpha}{(1+\alpha)(1+1/\alpha)}= 1 $ so $ \frac{1}{1+\sin^2x}+\frac{1}{1+\csc^2x} = 1 $ wherever $\csc x$ is defined and similarly $ \frac{1}{1+\cos^2x}+\frac{1}{1+\sec^2x} = 1 $