I have tried a lot of ways to solve this question but I am unable to get the answer as same as my textbook.
The text book answer is as follow: $$\frac{1}{2}\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)$$
The steps which I took is as follows:
$$\sqrt{\frac{1+ \sin x}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}}$$
Then Secondly
$$\sqrt{\frac{\left(1+\sin x\right)^2}{1-\sin^2 x}}$$
Then I got
$$\dfrac{1+\sin x}{\cos x}$$
When I differentiated this I got the following
$$\frac{\cos ^2\left(x\right)+\sin \left(x\right)\left(1+\sin \left(x\right)\right)}{\cos ^2\left(x\right)}$$
Can Anyone tell me what I am doing wrong?
I also know that $$\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)=\frac{2}{\left(\cos \frac{x}{2}-\sin\frac{x}{2}\right)^2}$$
Thank you for the help!
Best Answer
Why make things complicated if there is a easy way? By the half-angle formulae we obtain
$$\sqrt{\frac{1-\sin(x)}{1+\sin(x)}}=\sqrt{\frac{1-\cos\left(x+\frac\pi2\right)}{1+\cos\left(x+\frac\pi2\right)}}=\tan\left(\frac x2-\frac\pi4\right)$$
And I suppose you can differentiate the tangent function ;)
As pointed out by Simply Beautiful Art and mathcounterexamples.net by using the half-angle formula we ran into serious issus concerning the sign.