Question

If the area of triangle $ABC$ is $b^2 - (c-a)^2$, then $\tan B =$

• A. 1
• B. $\dfrac{13}{15}$
• C. $\dfrac{1}{4}$
• D. $\dfrac{8}{15}$

Hint:

Whenever an area is expressed in terms of sides, use the Cosine Rule to eliminate the side terms and reduce to a single trigonometric equation in one angle.

Answer 1

The Correct Option is D

Step 1: Concept
The area of a triangle is $\Delta = \tfrac{1}{2}ac\sin B$. Expand the given expression: $b^2-(c-a)^2 = b^2 - c^2 - a^2 + 2ac$.

Step 2: Meaning
By the Cosine Rule, $b^2 - c^2 - a^2 = -2ac\cos B$. So the given area equals $2ac(1-\cos B)$.

Step 3: Analysis
Equate the two expressions for area: \[\tfrac{1}{2}ac\sin B = 2ac(1-\cos B) \implies \sin B = 4(1-\cos B).\] Using half-angle substitution: $2\sin\tfrac{B}{2}\cos\tfrac{B}{2} = 8\sin^2\!\tfrac{B}{2} \implies \tan\tfrac{B}{2} = \tfrac{1}{4}$.

Step 4: Conclusion
\[\tan B = \frac{2\tan(B/2)}{1-\tan^2(B/2)} = \frac{2\cdot\tfrac{1}{4}}{1-\tfrac{1}{16}} = \frac{\tfrac{1}{2}}{\tfrac{15}{16}} = \frac{8}{15}.\] Final Answer: (D)