How do i prove that left side equals right?

Question

How do i prove that left side equals right?

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

Answer 1

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 2

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 3

Thanks for the help

Answer 4

To prove that the left side equals the right side of the given equation, we need to apply trigonometric identities and manipulate the expressions to obtain an equivalent form. Here's how you can do it:

Starting with the left side of the equation:
Cot(A + B)

Step 1:
Apply the addition formula for cotangent:
cot(A + B) = cot(A)cot(B) − 1 / (cot(A) + cot(B))

Step 2:
Rearrange the denominator:
1 / (cot(A) + cot(B)) = 1 / cot(A) * 1 / (1 + cot(A)/cot(B))

Step 3:
Use the identity cot(A) = 1/tan(A) to rewrite the expressions:
1 / cot(A) * 1 / (1 + cot(A)/cot(B)) = tan(A) * 1 / (1 + (1/tan(A)) * cot(B))

Step 4:
Simplify the expression:
tan(A) * 1 / (1 + (1/tan(A)) * cot(B)) = tan(A) / (1/tan(A) + cot(B))

Step 5:
Apply the identity cot(B) = 1/tan(B):
tan(A) / (1/tan(A) + cot(B)) = tan(A) / (1/tan(A) + 1/tan(B))

Step 6:
Combine the fractions with a common denominator:
tan(A) / (1/tan(A) + 1/tan(B)) = tan(A) / ((tan(B) + tan(A)) / (tan(A) * tan(B)))

Step 7:
Invert the denominator and multiply:
tan(A) / ((tan(B) + tan(A)) / (tan(A) * tan(B))) = tan(A) * (tan(A) * tan(B)) / (tan(B) + tan(A))

Step 8:
Simplify:
tan(A) * (tan(A) * tan(B)) / (tan(B) + tan(A)) = tan(A)^2 * tan(B) / (tan(B) + tan(A))

Step 9:
Apply the identity tan(A)^2 = 1 - cot(A)^2:
tan(A)^2 * tan(B) / (tan(B) + tan(A)) = (1 - cot(A)^2) * tan(B) / (tan(B) + tan(A))

Step 10:
Apply the identity cot(A) = 1/tan(A):
(1 - cot(A)^2) * tan(B) / (tan(B) + tan(A)) = (1 - 1/tan(A)^2) * tan(B) / (tan(B) + tan(A))

Step 11:
Simplify:
(1 - 1/tan(A)^2) * tan(B) / (tan(B) + tan(A)) = (tan(A)^2 - 1) * tan(B) / (tan(B) + tan(A))

Step 12:
Apply the identity tan(B) = 1/cot(B):
(tan(A)^2 - 1) * tan(B) / (tan(B) + tan(A)) = (tan(A)^2 - 1) * (1/cot(B)) / (1/cot(B) + tan(A))

Step 13:
Simplify:
(tan(A)^2 - 1) * (1/cot(B)) / (1/cot(B) + tan(A)) = (tan(A)^2 - 1) / (1 + cot(B) * tan(A))

Step 14:
Apply the identity cot(B) * tan(A) = 1:
(tan(A)^2 - 1) / (1 + cot(B) * tan(A)) = (tan(A)^2 - 1) / (1 + 1)
= (tan(A)^2 - 1) / 2

Therefore, Cot(A + B) = (tan(A)^2 - 1) / 2, which proves that the left side equals the right side of the equation.