Question:
In a triangle \(ABC\),
\[
\frac{a-b}{a+b}=
\]
In a triangle \(ABC\),
\[
\frac{a-b}{a+b}=
\]
Show Hint
In triangle problems involving side ratios like
\[
\frac{a-b}{a+b},
\]
first use sine rule and then apply sum-to-product identities.
Updated On: Jun 26, 2026
- \(\cot\left(\dfrac{A-B}{2}\right)\cot\dfrac{C}{2}\)
- \(\tan\left(\dfrac{A+B}{2}\right)\tan\dfrac{C}{2}\)
- \(\tan\left(\dfrac{A-B}{2}\right)\tan\dfrac{C}{2}\)
- \(\tan\left(\dfrac{A+B+C}{2}\right)\)
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The Correct Option is C
Solution and Explanation
Step 1: Use sine rule.
In triangle \(ABC\), by sine rule, \[ \frac{a}{\sin A}=\frac{b}{\sin B} \] So, \[ \frac{a-b}{a+b} = \frac{\sin A-\sin B}{\sin A+\sin B} \]
Step 2: Use sum-to-product identities.
We know that \[ \sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \] and \[ \sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \]
Step 3: Substitute these identities.
Therefore, \[ \frac{\sin A-\sin B}{\sin A+\sin B} = \frac{2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)} {2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)} \]
Step 4: Simplify the expression.
Cancelling \(2\), we get \[ = \frac{\cos\left(\frac{A+B}{2}\right)}{\sin\left(\frac{A+B}{2}\right)} \cdot \frac{\sin\left(\frac{A-B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)} \] \[ = \cot\left(\frac{A+B}{2}\right) \tan\left(\frac{A-B}{2}\right) \]
Step 5: Use angle sum of triangle.
Since \[ A+B+C=180^\circ, \] we have \[ \frac{A+B}{2}=\frac{180^\circ-C}{2} \] \[ \frac{A+B}{2}=90^\circ-\frac{C}{2} \]
Step 6: Convert cotangent term.
Now, \[ \cot\left(90^\circ-\frac{C}{2}\right)=\tan\frac{C}{2} \] Hence, \[ \frac{a-b}{a+b} = \tan\left(\frac{A-B}{2}\right)\tan\frac{C}{2} \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\tan\left(\frac{A-B}{2}\right)\tan\frac{C}{2}} \]
In triangle \(ABC\), by sine rule, \[ \frac{a}{\sin A}=\frac{b}{\sin B} \] So, \[ \frac{a-b}{a+b} = \frac{\sin A-\sin B}{\sin A+\sin B} \]
Step 2: Use sum-to-product identities.
We know that \[ \sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \] and \[ \sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \]
Step 3: Substitute these identities.
Therefore, \[ \frac{\sin A-\sin B}{\sin A+\sin B} = \frac{2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)} {2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)} \]
Step 4: Simplify the expression.
Cancelling \(2\), we get \[ = \frac{\cos\left(\frac{A+B}{2}\right)}{\sin\left(\frac{A+B}{2}\right)} \cdot \frac{\sin\left(\frac{A-B}{2}\right)}{\cos\left(\frac{A-B}{2}\right)} \] \[ = \cot\left(\frac{A+B}{2}\right) \tan\left(\frac{A-B}{2}\right) \]
Step 5: Use angle sum of triangle.
Since \[ A+B+C=180^\circ, \] we have \[ \frac{A+B}{2}=\frac{180^\circ-C}{2} \] \[ \frac{A+B}{2}=90^\circ-\frac{C}{2} \]
Step 6: Convert cotangent term.
Now, \[ \cot\left(90^\circ-\frac{C}{2}\right)=\tan\frac{C}{2} \] Hence, \[ \frac{a-b}{a+b} = \tan\left(\frac{A-B}{2}\right)\tan\frac{C}{2} \]
Step 7: Final conclusion.
Therefore, \[ \boxed{\tan\left(\frac{A-B}{2}\right)\tan\frac{C}{2}} \]
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