Question

How do i prove that left side equals right?

Cot(A+B)=cotAcotB-1/(cotA + cotB)

Answer 1

i wonder why that should be a problem
recat the inverse of tangent gives contangent......thus
1/tan(a+b)=tan(a)+tan(b)-1/tan(a+b/)
1/tan(a+b)=tan(a)+tan(b)-1/tan(a)+tan(b)
now divide through by tan(a)tan(b)
1/tan(a+b)=1/tan(b)+tan(a)/1/tan(a)+1/tan(b)
thus
cot(a+b)=cot(a)cot(b)-1/cot(a)cot(b)....if i did not make any typo that should be it

Answer 2

To prove an identity to be true you have to simplify the LS and the RS independently until you have LS = RS

The first thing I usually try is to take any angle and sub it into the equation to see if it is valid.
Here I took A=20, B=30
LS= cot(50) = .839...
RS = cot20cot30-1/(cot20+cot30)
not = to LS

but if we change it to
Cot(A+B)=(cotAcotB-1)/(cotA + cotB)
it does work

so:
Cot(A+B)=(cotAcotB-1)/(cotA + cotB)

RS = ((cosA/sinA)(cosB/sinB) -1)/(cosA/sinA + cosB/sinB)
= (cosAcosB/(sinAsinB)-sinAsinB/(sinAsinB)/(sinBcosA + cosBsinA)/(sinAsinB))
= (cosAcosB-sinAsinB)/(sinAsinB)*(sinAsinB)/(sinBcosA + cosBsinA)
= cos(A+B)/sin(A+B)
= cot(A+b)
= LS

Answer 3

Thanks for the help