Question

If in a $\triangle ABC$, with usual notations, $a^2$, $b^2$, $c^2$ are in A.P., then $\frac{\sin 3B}{\sin B} = $

• A. $\frac{a^2 - c^2}{2ac}$
• B. $\left(\frac{a^2 - c^2}{2ac}\right)^2$
• C. $\frac{a^2 - c^2}{ac}$
• D. $\left(\frac{a^2 - c^2}{ac}\right)^2$

Hint:

When an identity problem involves an A.P. condition, pick easy values to find the answer quickly! Let $a^2=1$, $b^2=25$, and $c^2=49$ (so $a=1, b=5, c=7$). These form a right-angled or valid triangle structure where you can compute the values directly to verify matching exponent powers inside the denominator!

Answer 1

The Correct Option is B

Step 1: Use the A.P. condition.
Since $a^2, b^2, c^2$ are in A.P., $2b^2 = a^2 + c^2$, so $b^2 = \dfrac{a^2 + c^2}{2}$.
Step 2: Expand the triple angle.
$\dfrac{\sin 3B}{\sin B} = \dfrac{3\sin B - 4\sin^3 B}{\sin B} = 3 - 4\sin^2 B = 4\cos^2 B - 1$.
Step 3: Bring in the Law of Cosines.
$\cos B = \dfrac{a^2 + c^2 - b^2}{2ac}$. Substituting $b^2 = \dfrac{a^2+c^2}{2}$ gives $\cos B = \dfrac{(a^2+c^2)/2}{2ac} = \dfrac{a^2 + c^2}{4ac}$.
Step 4: Square and scale.
$4\cos^2 B = 4\cdot\dfrac{(a^2+c^2)^2}{16a^2c^2} = \dfrac{(a^2+c^2)^2}{4a^2c^2}$.
Step 5: Subtract one over a common denominator.
$\dfrac{(a^2+c^2)^2 - 4a^2c^2}{4a^2c^2} = \dfrac{a^4 - 2a^2c^2 + c^4}{4a^2c^2} = \dfrac{(a^2-c^2)^2}{4a^2c^2}$.
Step 6: Write as a square.
This is $\left(\dfrac{a^2 - c^2}{2ac}\right)^2$. Recognising the numerator as a perfect square is the key simplification.
\[ \boxed{\left(\dfrac{a^2 - c^2}{2ac}\right)^2} \]