Answer: see proof below
Step-by-step explanation:
Given: A + B + C = π and cos A = cos B · cos C
scratchwork:
A + B + C = π
A = π - (B + C)
cos A = cos [π - (B + C)] Apply cos
= - cos (B + C) Simplify
= -(cos B · cos C - sin B · sin C) Sum Identity
= sin B · sin C - cos B · cos C Simplify
cos B · cos C = sin B · sin C - cos B · cos C Substitution
2cos B · cos C = sin B · sin C Addition
[tex]2=\dfrac{\sin B\cdot \sin C}{\cos B \cdot \cos C}[/tex] Division
2 = tan B · tan C
[tex]\text{Use the Sum Identity:}\quad \tan(B+C)=\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}[/tex]
Proof LHS → RHS
Given: A + B + C = π
Subtraction: A = π - (B + C)
Apply tan: tan A = tan(π - (B + C))
Simplify: = - tan (B + C)
[tex]\text{Sum Identity:}\qquad \qquad \qquad =-\bigg(\dfrac{\tan B+\tan C}{1-\tan B\cdot \tan C}\bigg)[/tex]
Substitution: = -(tan B + tan C)/(1 - 2)
Simplify: = -(tan B + tan C)/-1
= tan B + tan C
LHS = RHS: tan B + tan C = tan B + tan C [tex]\checkmark[/tex]
