Question
How do i prove that left side equals right?
Cot(A+B)=cotAcotB-1/(cotA + cotB)
Cot(A+B)=cotAcotB-1/(cotA + cotB)
Answers
collins
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
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
Reiny
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
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
Calista
Thanks for the help