Got one question from book, i try to do but my answer is wrong...
Differentiate the function with respect to A.
In(cosecA + cotA)
After differentiate cosecA+cotA i stuck.
Answer: -cosecA
First of all, it's $\displaystyle \ln(\csc{A} + \cot{A})$, which stands for Natural Logarithm. Write the inverse trig functions in terms of what you already know...
So $\displaystyle y = \ln(\frac{1}{\sin{A}} + \frac{\cos{A}}{\sin{A}}) = \ln(\frac{1 + \cos{A}}{\sin{A}})$.
You have a composition of functions, so you need to use the chain rule.
Let $\displaystyle u = \frac{1 + \cos{A}}{\sin{A}}$, so $\displaystyle y = \ln{u}$.
Can you go from there?
Note, to find $\displaystyle \frac{du}{dx}$ you'll have to use the Quotient rule.
I transform the cosec-function and the cot-function into:
$\displaystyle cosec(A)=\dfrac1{\sin(A)}$ and $\displaystyle \cot(A)=\dfrac1{\tan(A)}$ Then your term becomes:
$\displaystyle f(A) = \ln\left(\dfrac1{\sin(A)} + \dfrac1{\tan(A)}\right)$ Use chain rule to differentiate:
$\displaystyle f'(A) = \dfrac1{\dfrac1{\sin(A)} + \dfrac1{\tan(A)}} \cdot \left(-\dfrac1{\sin^2(A)} \cdot \cos(A) - \dfrac1{\sin^2(A)}\right)$
$\displaystyle f'(A)=\dfrac{\sin(A) \cdot \tan(A)}{\sin(A) + \tan(A)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)}$ ........ Transform $\displaystyle \tan(A)=\dfrac{\sin(A)}{\cos(A)}$ which will yield:
$\displaystyle f'(A)=\dfrac{\sin^2(A)}{\cos(A) \cdot \left(\sin(A)+\dfrac{\sin(A)}{\cos(A)}\right)} \cdot \dfrac{-(\cos(A)+1)}{\sin^2(A)}$ ...... Expand the bracket and cancel out $\displaystyle \sin^2(A)$
$\displaystyle f'(A)=\dfrac1{\sin(A) \cdot \cos(A) + \sin(A)} \cdot \dfrac{-(\cos(A)+1)}{1}$
$\displaystyle f'(A)=\dfrac1{\sin(A) \cdot (\cos(A) + 1)} \cdot \dfrac{-(\cos(A)+1)}{1}= -\dfrac1{\sin(A)} = - cosec(A)$
Heavy weather, earboth .....
Continuing on from Life might be easier going one step further:
$\displaystyle \frac{dy}{dA} = \frac{-\sin A}{1 + \cos A} - \frac{\cos A}{\sin A} = \frac{-\sin^2 A - \cos A - \cos^2 A}{\sin A (1 + \cos A)}$ $\displaystyle = \frac{-1 - \cos A}{\sin A (1 + \cos A)} = \frac{-1}{\sin A} = - \text{cosec} A$.