Question

If in $\Delta ABC$, with usual notations, the angles are in A.P., then $\frac{a}{c}\sin(2C) + \frac{c}{a}\sin(2A) =$

• A. $\frac{\sqrt{3}}{2}$
• B. $3$
• C. $2\sqrt{3}$
• D. $\sqrt{3}$

Hint:

When an expression depends generally on properties of an A.P. triangle, you can choose a convenient specific case to find the answer instantly! Let $\Delta ABC$ be an equilateral triangle where $A = B = C = 60^\circ$ (which satisfies the A.P. condition) and $a = b = c$. The expression simplifies to:
$$\frac{a}{a}\sin(120^\circ) + \frac{a}{a}\sin(120^\circ) = 2\sin(120^\circ) = 2 \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}$$ This approach saves a lot of algebraic steps during competitive exams.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a triangle $ABC$ whose interior angles follow an Arithmetic Progression (A.P.). We need to evaluate the scalar value of the given trigonometric side-ratio expression.

Step 2: Key Formula or Approach:
1. Since the angles are in A.P. and $A + B + C = 180^\circ$, the middle angle must be $B = 60^\circ$. This leaves $A + C = 120^\circ$. 2. Use the Sine Rule to relate sides and angles: $\frac{a}{\sin A} = \frac{c}{\sin C} \implies a = \lambda\sin A$ and $c = \lambda\sin C$. 3. Substitute these expressions into the target relation and simplify using double-angle and compound-angle formulas: $\sin(2\theta) = 2\sin\theta\cos\theta$.

Step 3: Detailed Explanation:
Let's substitute the Sine Rule relations $a = \lambda\sin A$ and $c = \lambda\sin C$ into the expression: $$E = \frac{\lambda\sin A}{\lambda\sin C}(2\sin C\cos C) + \frac{\lambda\sin C}{\lambda\sin A}(2\sin A\cos A)$$ The $\lambda$ factors and the matching denominator sine terms cancel out cleanly: $$E = 2\sin A\cos C + 2\sin C\cos A$$ Factor out the constant 2: $$E = 2(\sin A\cos C + \cos A\sin C)$$ Recognize the trigonometric compound angle identity $\sin(A + C) = \sin A\cos C + \cos A\sin C$: $$E = 2\sin(A + C)$$ Since the angles are in A.P., we know $B = 60^\circ$, which means: $$A + B + C = 180^\circ \implies A + C = 180^\circ - 60^\circ = 120^\circ$$ Substitute $A + C = 120^\circ$ into the simplified expression: $$E = 2\sin(120^\circ) = 2\sin(180^\circ - 60^\circ) = 2\sin(60^\circ)$$ $$E = 2 \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}$$

Step 4: Final Answer:
The simplified value of the expression is $\sqrt{3}$, matching option (D).