NIMCET 2007 — Mathematics PYQ

To solve this problem, we will use the half-angle formulas from the properties of a triangle alongside the condition for terms being in an Arithmetic Progression (A.P.).

Let $s$ be the semi-perimeter of the triangle $\Delta ABC$, defined as:

$$s=\frac{a+b+c}{2}$$

Step 1: Apply the given A.P. condition

Since the sides $a$, $b$, and $c$ are in Arithmetic Progression (A.P.):

$$2b=a+c$$

Now, substitute this condition into the expression for the semi-perimeter:

$$2s=(a+c)+b$$

$$2s=2b+b$$

$$2s=3b⟹s=\frac{3b}{2}$$

Step 2: Use the half-angle formulas for tangents

From the standard properties of triangles, the half-angle trigonometric formulas for $\tan \frac{A}{2}$ and $\tan \frac{C}{2}$ are:

$$\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$$

$$\tan \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$

Step 3: Multiply the two tangent expressions

Now, let's find the value of the product $\tan \frac{A}{2}\cdot \tan \frac{C}{2}$:

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\cdot \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$

Combine both fractions under a single square root:

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=\sqrt{\frac{(s-b)(s-c)(s-a)(s-b)}{s(s-a)s(s-c)}}$$

Cancel out the common factors ($(s-a)$ and $(s-c)$) from the numerator and denominator:

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=\sqrt{\frac{(s-b{)}^2}{s^2}}$$

Simplifying the square root gives:

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=\frac{s-b}{s}$$

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=1-\frac{b}{s}$$

Step 4: Substitute the value of $s$

From Step 1, we established that $s=\frac{3b}{2}$. Substituting this value back into our simplified expression:

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=1-\frac{b}{\left(\frac{3b}{2}\right)}$$

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=1-\frac{2b}{3b}$$

The variable $b$ cancels out:

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=1-\frac{2}{3}$$

$$\tan \frac{A}{2}\cdot \tan \frac{C}{2}=\frac{1}{3}$$

Correct Answer

The correct option is (a) $\frac{1}{3}$.