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Find the derivative with respect to $x$ of $$\arctan \left( \cos x\over1+\sin x \right).$$

I tried solving this problem by using trigonometric functions of submultiple of numbers (formulas of $\sin x$ and $\cos x$) but they didn't help.

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Hint:

$$\frac{\cos x}{1+\sin x} = \frac{\sin(\frac\pi2 -\frac x2 )}{1+\cos(\frac\pi2-\frac x2) } = \frac{2\sin(\frac\pi4-\frac x2)\cos(\frac\pi4-\frac x2)}{2\cos^2(\frac\pi4-\frac x2)} = \tan(\frac\pi4-\frac x2).$$

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  • $\begingroup$ Thabks i got the answer its 1/2 $\endgroup$
    – Mad Dawg
    Jul 26 '19 at 17:19
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    $\begingroup$ You're welcome! But it should be $-\frac 12$ $\endgroup$
    – Ak.
    Jul 26 '19 at 17:20
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    $\begingroup$ Oh yeah yeah right its - 1/2 $\endgroup$
    – Mad Dawg
    Jul 26 '19 at 17:20
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It's $$\frac{1}{1+\left(\frac{\cos{x}}{1+\sin{x}}\right)^2}\cdot\left(\frac{\cos{x}}{1+\sin{x}}\right)'.$$ Can you end it now?

Also, you can use $$\frac{\cos{x}}{1+\sin{x}}=\frac{\sin\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right)}=\tan\left(\frac{\pi}{4}-\frac{x}{2}\right).$$

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    $\begingroup$ You got it, I missed $\frac x2$ and typed it as $x$. + 1 $\endgroup$
    – Ak.
    Jul 26 '19 at 14:39
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Using https://www.cut-the-knot.org/arithmetic/algebra/WeierstrassSubstitution.shtml

$y=\arctan\dfrac{\cos x}{1+\sin x}=\arctan\dfrac{1-t^2}{(1+t)^2}$ where $t=\tan\dfrac x2$

If $1+t\ne0$

$y=n\pi+\dfrac\pi4-\dfrac x2$ where $n$ is an integer such that $-\dfrac\pi2<y<\dfrac\pi2$

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