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Given $$\cot{\theta} = \frac{\sqrt{1+\sin(x)}+\sqrt{1-\sin(x)}}{\sqrt{1+\sin(x)}-\sqrt{1-\sin(x)}}$$ I have to find its differential coefficient w.r.t $x$ i.e. $\dfrac{d \theta }{dx}$.

Now I can find it in the following two ways:

(1). When I write $\sqrt{1-\sin(x)}=\cos(x/2)-\sin(x/2)$, I get $$\cot(\theta) = \frac{\cos(x/2) + \sin(x/2) + \cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2) - \cos(x/2) + \sin(x/2)} = \cot(x/2)$$ $$\theta=x/2 \implies \frac{d \theta}{dx} = \frac12$$

(2). When I write $\sqrt{1-\sin(x)}=\sin(x/2)-\cos(x/2)$, I get $$\cot(\theta) = \frac{\cos(x/2) + \sin(x/2) - \cos(x/2) + \sin(x/2)}{\cos(x/2) + \sin(x/2) + \cos(x/2) - \sin(x/2)} = \tan(x/2)$$ $$\theta=\pi/2 - x/2 \implies \frac{d \theta}{dx} = -\frac12$$ Which of the above two answers is correct and why? Please help me know it. Thanks.

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Dear Sanjeev Verma, welcome to math stackexchange. Kindly refer here (…) on how to typeset (i.e. how to write equations etc) the question. Currently, it is very hard to read what you have typed out. – user17762 May 20 '12 at 5:41
Very well done @Marvis. – Gigili May 20 '12 at 6:00
Thanks @Gigili. – user17762 May 20 '12 at 6:15
thank you very much Marvis for your help and also for giving the above link – Sanjeev Verma May 20 '12 at 8:36

Neither. When you deal in half-angle formulas, you have to be careful about sign.

Notice that $\sqrt{1-\sin x}$ is always positive, while $\sin (x/2)-\cos(x/2)$ and $\cos(x/2)-\sin(x/2)$ may be positive or negative depending on $x$. The correct statement is $\sqrt{1-\sin x}=|\cos(x/2)-\sin(x/2)|$; from there, you'll need to do some case analysis to compute the derivative you want.

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Thanks for your answer Micah – Sanjeev Verma May 20 '12 at 8:23

Micah has a good answer as to why you have an inconsistency. To make your task easier, consider rationalizing your denominator:

$$\begin{align} \frac{\sqrt{1+\sin(x)}+\sqrt{1-\sin(x)}}{\sqrt{1+\sin(x)}-\sqrt{1-\sin(x)}}\cdot\frac{\sqrt{1+\sin(x)}+\sqrt{1-\sin(x)}}{\sqrt{1+\sin(x)}+\sqrt{1-\sin(x)}}&=\frac{2+2\sqrt{1-\sin^2(x)}}{2\sin(x)}\\ \cot(\theta)&=\frac{1+|\cos(x)|}{\sin(x)} \end{align}$$

Now the chain rule, quotient rule, and the derivative of $|x|$ (which is $|x|/x$) will give you $\frac{d\theta}{dx}$ in terms of $x$ and $\csc^2(\theta)$ without dealing with any more square roots or half-angle formulas. If you like, $\csc^2(\theta)$ can be subbed out for $1+\cot^2(\theta)$, which can be written in terms of $x$ alone.

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Thanks to you too alex.jordan – Sanjeev Verma May 20 '12 at 8:24

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