Question

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

• A. \(1\)
• B. \(13/15\)
• C. \(1/4\)
• D. \(8/15\)

Hint:

Whenever expressions like \(1-\cos\theta\) appear in triangle problems, immediately convert them using the half-angle identity: \[ 1-\cos\theta = 2\sin^2\frac{\theta}{2} \] This transformation is extremely useful in simplifying area and trigonometric relation questions.

Answer 1

The Correct Option is D

Step 1: Expanding the given area expression carefully.
The area of the triangle is given as: \[ \Delta = b^2 - (c-a)^2 \] Expand the square term: \[ (c-a)^2 = c^2 + a^2 - 2ac \] Therefore, \[ \Delta = b^2 - c^2 - a^2 + 2ac \] Now apply the cosine rule in \(\triangle ABC\): \[ b^2 = a^2 + c^2 - 2ac\cos B \] Substituting this value of \(b^2\) into the area expression: \[ \Delta = (a^2+c^2-2ac\cos B)-c^2-a^2+2ac \] Simplifying: \[ \Delta = -2ac\cos B + 2ac \] \[ \Delta = 2ac(1-\cos B) \] Using the identity: \[ 1-\cos B = 2\sin^2\frac{B}{2} \] Hence, \[ \Delta = 2ac\left(2\sin^2\frac{B}{2}\right) \] \[ \Delta = 4ac\sin^2\frac{B}{2} \]

Step 2: Using the standard area formula of a triangle.
We know that the area of a triangle can also be written as: \[ \Delta = \frac12 ac\sin B \] Using: \[ \sin B = 2\sin\frac{B}{2}\cos\frac{B}{2} \] we get: \[ \Delta = \frac12 ac \left(2\sin\frac{B}{2}\cos\frac{B}{2}\right) \] Thus, \[ \Delta = ac\sin\frac{B}{2}\cos\frac{B}{2} \] Now equate both expressions of area: \[ ac\sin\frac{B}{2}\cos\frac{B}{2} = 4ac\sin^2\frac{B}{2} \] Cancel \(ac\) from both sides: \[ \sin\frac{B}{2}\cos\frac{B}{2} = 4\sin^2\frac{B}{2} \] Divide by \(\sin\frac{B}{2}\): \[ \cos\frac{B}{2} = 4\sin\frac{B}{2} \] Therefore, \[ \tan\frac{B}{2}=\frac14 \]

Step 3: Applying the double-angle formula for tangent.
Use the identity: \[ \tan B = \frac{2\tan(B/2)}{1-\tan^2(B/2)} \] Substitute: \[ \tan\frac{B}{2}=\frac14 \] Then, \[ \tan B = \frac{2\left(\frac14\right)} {1-\left(\frac14\right)^2} \] \[ = \frac{\frac12}{1-\frac1{16}} \] \[ = \frac{\frac12}{\frac{15}{16}} \] \[ = \frac12 \times \frac{16}{15} \] \[ = \frac{8}{15} \] Hence, \[ \boxed{\tan B=\frac{8}{15}} \]