Verify: #-(cotA+cotB)/(cotA-cotB) = sin(A+B)/sin(A-B)# ?

Question

Anna Allen · Accepted Answer

Expand and simplify RHS then apply identities for #sin(A+-B)# RHS =#-(cotA+cotB)/(cotA-cotB) = -(cosA/sinA+cosB/sinB)/(cosA/sinA-cosB/sinB)# RHS=#-(cosAsinB + cosBsinA)/(sinAsinB)/(cosAsinB-cosBsinA)/(sinAsinB)# RHS=#-(cosAsinB + cosBsinA)/(cancel(sinAsinB))/(cosAsinB-cosBsinA)/(cancel(sinAsinB))# RHS=#-(cosAsinB + cosBsinA)/(cosAsinB-cosBsinA)# #=-sin(A+B)/-sin(A-B) = sin(A+B)/sin(A-B) =# LHS

Serenity Johnson · Answer

#LHS=sin(A+B)/sin(A-B)# #=(sinAcosB+cosAsinB)/(sinAcosB-cosAsinB)# Dividing both numerator and denominator by #sinAsinB# #=((sinAcosB)/(SinAsinB)+(cosAsinB)/(sinAsinB))/((sinAcosB)/(sinAsinB)-(cosAsinB)/(sinAsinB))# #=-((cotA+cotB)/(cotA-cotB))=RHS# Verified

Anna Allen · Answer

undefined To verify the equation $ -\frac{\cot A + \cot B}{\cot A - \cot B} = \frac{\sin(A + B)}{\sin(A - B)} $, we'll start by expressing cotangents in terms of sines and cosines:

$$ \cot A = \frac{\cos A}{\sin A} $$
$$ \cot B = \frac{\cos B}{\sin B} $$

Substitute these expressions into the left-hand side of the equation:

$$ -\frac{\frac{\cos A}{\sin A} + \frac{\cos B}{\sin B}}{\frac{\cos A}{\sin A} - \frac{\cos B}{\sin B}} $$

Simplify this expression by finding a common denominator:

$$ -\frac{(\cos A \sin B + \cos B \sin A)}{(\cos A \sin B - \cos B \sin A)} $$

Now, use the sum and difference identities for sine:

$$ \sin(A + B) = \sin A \cos B + \cos A \sin B $$
$$ \sin(A - B) = \sin A \cos B - \cos A \sin B $$

Substitute these identities into the expression:

$$ -\frac{\sin(A + B)}{\sin(A - B)} $$

This matches the right-hand side of the equation. Therefore, the equation is verified.