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

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.

Hint:

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

• Thabks i got the answer its 1/2 Jul 26 '19 at 17:19
• You're welcome! But it should be $-\frac 12$
– Ak.
Jul 26 '19 at 17:20
• Oh yeah yeah right its - 1/2 Jul 26 '19 at 17:20

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

• You got it, I missed $\frac x2$ and typed it as $x$. + 1
– Ak.
Jul 26 '19 at 14:39

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