Easy STEP Question

OK, I'm going to kick myself after this but I can't seem to get started on this question, once I've done this part I'm sure I can do the rest. Anyway,

"The positive numbers, a, b and c satisy bc = a^2 + 1. Prove that

arctan (1/(a+b)) + arctan (1/(a+c)) = arctan(1/a)
"

How do you get rid of the arctans? Do you even need to? You can't just remove them can you? Anyway, thanks for any help, rep will be given!

Reply 1

7589200

maybe try drawing a triangle?

sides: root bc, a and 1.

Reply 2

bob_54321

ok, thanks a lot. i always forget to draw diagrams and tend to try to do things by algebra manipulation. really not a good habit, lol.

Reply 3

evariste_3086

bob_54321

let A=arctan (1/(a+b)) B=arctan (1/(a+c))

then Tan(A+B)=1/(a+b)+1/(a+c)/1-1/(a+b)(a+c)
=(2a+b+c)/(a^2+a(b+c)+bc-1

=(2a+b+c)/2a^2+a(b+c) ..........using bc=a^2+1
=1/a[(2a+b+c)/(2a+b+c)]
=1/a
hence
A+B=arctan(1/a)

Reply 4

C4>O7

|
| \
|a \√(bc)
|__ \
__1

Having drawn this triangle I can't see how to show 'arctan....+.... = arctan(1/a)' :s-smilie:

I did it algebraically by 'taning' both sides.

Reply 5

bob_54321

yeah, i did the triangle thing but it didnt seem obvious how to get the result. are you meant to split the angle or something? i prefer evariste's method, thanks.