Inverse Trig functions- Find sin(arctanp+arctan1/p) if p>0
Hi, I was given this question for homework but I'm stuck. Determine sin(arctanp+arctan1/p) if p>0
After using the compound angle formula and substituting functions, I have no idea what to do next:
I started off by using the compound angle formula, giving me: [sin(arctan(p))]*[cos(arctan(1/p)] + [cos(arctan(p)]*[sin(arctan(1/p)]. From there I Let arctan(p)=a. Through manipulation I therefore got tan(arctan(p))=tan(a), and so p=tan(a)...I did the same for arctan(1/p) but with b, giving me (1/p)=tan(b)...
I realize that the answer to this question should be 1 but have no idea how to get it. Any help would be greatly appreciated! Thanks! :)