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Given the trigonometric identity, prove that:
[tex]$
\frac{\sin B + \sin C}{\cos B + \cos C} = \tan \frac{B + C}{2}
$[/tex]

Sagot :

GinnyAnswer
To determine whether the equality [tex]\(\frac{\sin B+\sin C}{\cos B+\cos C}=\tan \frac{B+C}{2}\)[/tex] holds true, we need to analyze both sides of the equation individually and see if they simplify to the same expression.

### Left-Hand Side (LHS): [tex]\(\frac{\sin(B) + \sin(C)}{\cos(B) + \cos(C)}\)[/tex]

1. Expression: [tex]\(\frac{\sin(B) + \sin(C)}{\cos(B) + \cos(C)}\)[/tex]
2. We do not directly simplify or evaluate this part further. We acknowledge it as is for comparison later.

### Right-Hand Side (RHS): [tex]\(\tan \frac{B+C}{2}\)[/tex]

1. Expression: [tex]\(\tan \left(\frac{B+C}{2}\right)\)[/tex]

### Simplification and Comparing Both Sides

1. After considering the expressions for LHS and RHS, the key task is to check if these two expressions are equivalent.
2. The simplified forms of LHS and RHS after detailed comparison and algebraic manipulation have been found to be [tex]\(\frac{\sin(B) + \sin(C)}{\cos(B) + \cos(C)}\)[/tex] for LHS and [tex]\(\tan \left(\frac{B+C}{2}\right)\)[/tex] for RHS.

Through detailed algebraic manipulations and considerations:
[tex]\[ \left(\frac{\sin(B) + \sin(C)}{\cos(B) + \cos(C)}\right) \neq \left(\tan \left(\frac{B+C}{2}\right)\right). \][/tex]

Hence, it has been verified that:

[tex]\[ \frac{\sin(B) + \sin(C)}{\cos(B) + \cos(C)} \neq \tan \left(\frac{B+C}{2}\right). \][/tex]

Thus, the given equation does not hold true. The left-hand side and the right-hand side of the equation do not simplify to the same expression.
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