Need some help with trigonometry. Inverse functions.

Question

Can someone help me to solve these examples?

Evaluate:

$$A= \text{arccot}\bigl(\tan\biggl[\frac{4\pi}{5}\biggr]\bigr)\;\; \text{and}\;\; B=\sin\bigl(\arctan\biggr[\frac{-3}{7}\biggr]\bigr)$$

Answer 1

I just signed up StackExchange today, and this is my first answer. Sorry, cause I can't use MathJax code properly. I will solve A first...
If $0\le A\le2\pi$, then there are 2 solution for A

$cotA = tan{(4\pi/5)}$
$cosA/sinA = sin(4\pi/5)/cos(4\pi/5)$
$cosA.cos(4\pi/5) = sinA.sin(4\pi/5)$
** Use trigonometry multiplication formula **
$1/2(cos(A+4\pi/5)+cos(A-4\pi/5))=-1/2(cos(A+4\pi/5)-cos(A-4\pi/5))$
$cos(A+4\pi/5)+cos(A-4\pi/5)=-cos(A+4\pi/5)+cos(A-4\pi/5)$
$cos(A+4\pi/5)=-cos(A+4\pi/5)$
$2cos(A+4\pi/5)=0$
$cos(A+4\pi/5)=0$
$cos(A+4\pi/5)=cos(\pi/2)$
$A+4\pi/5=\pi/2$
$A=-3\pi/10=17\pi/10$, this is the first solution.

Now back to $cos(A+4\pi/5)=0$,
because not only $cos(\pi/2)=0$,
$cos(3\pi/2)$ also equals $0$, so we should substitute $cos(3\pi/2)$ to $0$ too...

$cos(A+4\pi/5)=cos(3\pi/2)$
$A+4\pi/5=3\pi/2$
$A=7\pi/10$, this is the second solution
Hope this help