Question:
In any triangle \(ABC\),
\[
a(b\cos C-c\cos B)=
\]
In any triangle \(ABC\),
\[
a(b\cos C-c\cos B)=
\]
Show Hint
For expressions involving sides and cosines in a triangle, use the cosine rule to convert trigonometric terms into side lengths.
Updated On: Jun 25, 2026
- \(b-c\)
- \(b+c\)
- \(b^2-c^2\)
- \(b^2+c^2\)
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The Correct Option is C
Solution and Explanation
Step 1: Use cosine rule for \(\cos C\).
In triangle \(ABC\), by cosine rule, \[ c^2=a^2+b^2-2ab\cos C \] Therefore, \[ 2ab\cos C=a^2+b^2-c^2 \] So, \[ \cos C=\frac{a^2+b^2-c^2}{2ab} \]
Step 2: Use cosine rule for \(\cos B\).
Again, by cosine rule, \[ b^2=a^2+c^2-2ac\cos B \] Therefore, \[ 2ac\cos B=a^2+c^2-b^2 \] So, \[ \cos B=\frac{a^2+c^2-b^2}{2ac} \]
Step 3: Substitute in the given expression.
We need to simplify \[ a(b\cos C-c\cos B) \] First find \[ b\cos C-c\cos B \] Substitute the values: \[ b\cos C-c\cos B = b\left(\frac{a^2+b^2-c^2}{2ab}\right) - c\left(\frac{a^2+c^2-b^2}{2ac}\right) \] \[ = \frac{a^2+b^2-c^2}{2a} - \frac{a^2+c^2-b^2}{2a} \]
Step 4: Simplify the expression.
\[ b\cos C-c\cos B = \frac{a^2+b^2-c^2-a^2-c^2+b^2}{2a} \] \[ = \frac{2b^2-2c^2}{2a} \] \[ = \frac{b^2-c^2}{a} \] Now multiply by \(a\): \[ a(b\cos C-c\cos B) = a\cdot \frac{b^2-c^2}{a} \] \[ =b^2-c^2 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{b^2-c^2} \]
In triangle \(ABC\), by cosine rule, \[ c^2=a^2+b^2-2ab\cos C \] Therefore, \[ 2ab\cos C=a^2+b^2-c^2 \] So, \[ \cos C=\frac{a^2+b^2-c^2}{2ab} \]
Step 2: Use cosine rule for \(\cos B\).
Again, by cosine rule, \[ b^2=a^2+c^2-2ac\cos B \] Therefore, \[ 2ac\cos B=a^2+c^2-b^2 \] So, \[ \cos B=\frac{a^2+c^2-b^2}{2ac} \]
Step 3: Substitute in the given expression.
We need to simplify \[ a(b\cos C-c\cos B) \] First find \[ b\cos C-c\cos B \] Substitute the values: \[ b\cos C-c\cos B = b\left(\frac{a^2+b^2-c^2}{2ab}\right) - c\left(\frac{a^2+c^2-b^2}{2ac}\right) \] \[ = \frac{a^2+b^2-c^2}{2a} - \frac{a^2+c^2-b^2}{2a} \]
Step 4: Simplify the expression.
\[ b\cos C-c\cos B = \frac{a^2+b^2-c^2-a^2-c^2+b^2}{2a} \] \[ = \frac{2b^2-2c^2}{2a} \] \[ = \frac{b^2-c^2}{a} \] Now multiply by \(a\): \[ a(b\cos C-c\cos B) = a\cdot \frac{b^2-c^2}{a} \] \[ =b^2-c^2 \]
Step 5: Final conclusion.
Therefore, \[ \boxed{b^2-c^2} \]
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