# A sum of terms like $(1 + \sin^2x)^{-1}$

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.

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Please be clearer about the equation. Some would read 1/2(1+sin^2 x) as $\frac 1{2(1+\sin^2 x)}$ or at least worry that you meant that. Either (1+sin^2 x)/2, (1/2)(1+sin^2 x), or (best) $\LaTeX$. Trying to solve the problem, I see I had it wrong, so please 1/(2(1+sin^2 x)), or $\LaTeX$ – Ross Millikan Jul 24 '12 at 14:23
I 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? – Ross Millikan Jul 24 '12 at 14:30

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.
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 $$\frac{1}{1+\cos^2x}+\frac{1}{1+\sec^2x} = 1$$