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Question

In a triangle \(ABC\), if \(a\), \(b\) and \(c\) are the lengths of the sides opposite to the angles \(A\), \(B\) and \(C\) respectively, then what is \(\frac{\sin(A-B)}{\sin(A+B)}\) equal to ?

This question was previously asked in
NDA 1 2026 GAT Question Paper (12-Apr-2026)
The correct answer is
\(\frac{a^2-b^2}{c^2}\)

Triangle Trigonometry: Evaluating \(\frac{\sin(A-B)}{\sin(A+B)}\)

We need to find the value of the expression \(\frac{\sin(A-B)}{\sin(A+B)}\) in a triangle \(ABC\), where \(a\), \(b\), and \(c\) are the side lengths opposite angles \(A\), \(B\), and \(C\) respectively.

Step 1: Use Triangle Angle Properties

In any triangle \(ABC\), the sum of angles is \(A+B+C = \pi\). Therefore, \(A+B = \pi - C\).

Using the sine property \(\sin(\pi - x) = \sin x\), we get:

\( \sin(A+B) = \sin(\pi - C) = \sin C \)

Step 2: Substitute into the Expression

Substitute \(\sin(A+B) = \sin C\) into the given expression:

\( \frac{\sin(A-B)}{\sin(A+B)} = \frac{\sin(A-B)}{\sin C} \)

Step 3: Apply Sine Expansion Formula

Use the sine subtraction formula \(\sin(A-B) = \sin A \cos B - \cos A \sin B\):

\( \frac{\sin(A-B)}{\sin C} = \frac{\sin A \cos B - \cos A \sin B}{\sin C} \)

Step 4: Apply the Sine Rule

The Sine Rule states \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\), where \(R\) is the circumradius. From this, we have \(\sin A = \frac{a}{2R}\), \(\sin B = \frac{b}{2R}\), and \(\sin C = \frac{c}{2R}\).

Substitute these into the expression:

\( \frac{\frac{a}{2R} \cos B - \cos A \frac{b}{2R}}{\frac{c}{2R}} \)

Multiply the numerator and denominator by \(2R\):

\( \frac{a \cos B - b \cos A}{c} \)

Step 5: Apply the Law of Cosines

The Law of Cosines gives:

  • \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\)
  • \(\cos B = \frac{a^2 + c^2 - b^2}{2ac}\)

Substitute these cosine values into the expression \(\frac{a \cos B - b \cos A}{c}\):

\( \frac{a \left( \frac{a^2 + c^2 - b^2}{2ac} \right) - b \left( \frac{b^2 + c^2 - a^2}{2bc} \right)}{c} \)

Simplify the numerator:

\( \frac{\frac{a^2 + c^2 - b^2}{2c} - \frac{b^2 + c^2 - a^2}{2c}}{c} \)

\( \frac{1}{c} \left( \frac{(a^2 + c^2 - b^2) - (b^2 + c^2 - a^2)}{2c} \right) \)

\( \frac{1}{2c^2} (a^2 + c^2 - b^2 - b^2 - c^2 + a^2) \)

\( \frac{1}{2c^2} (2a^2 - 2b^2) \)

\( \frac{2(a^2 - b^2)}{2c^2} \)

\( \frac{a^2 - b^2}{c^2} \)

Conclusion

The expression \(\frac{\sin(A-B)}{\sin(A+B)}\) simplifies to \(\frac{a^2 - b^2}{c^2}\). This matches Option C.

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Similar Questions

  1. In a triangle ABC, a = (1 + √3) cm, b = 2 cm and angle C = 60°, then the other two angles are

  2. Consider the following statements :

    1. ABC is right angled triangle

    2. The angles of the triangle are in AP

    Which of the statements given above is/are correct ?

  3. If c = 8, what is the area of the triangle ?

  4. What is the value of a + b + √2 c equal to ?

  5. What is the ratio of a2 ∶ b2 ∶ c2 ?

  6. Consider the following statements:

    1. If ABC is a right-angled triangle, right-angled at A, and if sin \(\rm B = \frac 1 3,\)  then cosec C = 3.

    2. If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles.

    Which of the above statements is/are correct?

  7. If the angles of a triangle ABC are in AP and b : c = √3 : √2, then what is the measure of angle A?

  8. In a triangle ABC if a = 2, b = 3 and sin A = 2/3, then what is angle B equal to?

  9. In a triangle ABC, sin A - cos B - cos C = 0. What is angle B equal to?

  10. The sides of a triangle are m, n and \(\rm \sqrt{m^2+n^2+mn}\) . What is the sum of the acute angles of the triangle?


Important Questions from Properties of Triangles

  1. In a triangle ABC, a = (1 + √3) cm, b = 2 cm and angle C = 60°, then the other two angles are

  2. Which of the following measures can form a triangle?

  3. Which of the following cannot be the sides of a triangle?

  4. If in a triangle ABC, \(\frac{{2\cos A}}{a} + \frac{{\cos B}}{b} + \frac{{2\cos C}}{c} = \frac{a}{{bc}} + \frac{b}{{ca}}\) then the value of the angle A is

  5. If the data given to construct a triangle ABC are a = 5, b = 7, \(\sin A = \frac{3}{4}\), then it is possible to construct

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