Question

In any $\Delta ABC$, $\tan \frac{B+C}{2} =$

• A. $c \cot \frac{A}{2}$
• B. $\cot \frac{A}{2}$
• C. $\tan \frac{A}{2}$
• D. $\tan \frac{C}{2}$

Hint:

This rule applies to all trigonometric ratios in a triangle: sine becomes cosine, tangent becomes cotangent, and secant becomes cosecant when dealing with the half-angle of the remaining sum. For example, $\sin \frac{A+B}{2} = \cos \frac{C}{2}$.

Answer 1

The Correct Option is B

This problem utilizes the angle sum property of a triangle combined with trigonometric co-function identities. Step 1: Use the Angle Sum Property In any triangle $ABC$, the sum of the interior angles is always $180^\circ$ (or $\pi$ radians): $$A + B + C = 180^\circ$$

Step 2: Isolate the required angle sum We need to find an expression for $\frac{B+C}{2}$. First, isolate $(B+C)$: $$B + C = 180^\circ - A$$ Now, divide the entire equation by 2: $$\frac{B+C}{2} = \frac{180^\circ - A}{2}$$ $$\frac{B+C}{2} = 90^\circ - \frac{A}{2}$$

Step 3: Apply the Tangent Function Take the tangent of both sides: $$\tan\left(\frac{B+C}{2}\right) = \tan\left(90^\circ - \frac{A}{2}\right)$$

Step 4: Use Co-function Identity According to the trigonometric identity $\tan(90^\circ - \theta) = \cot \theta$: $$\tan\left(90^\circ - \frac{A}{2}\right) = \cot \frac{A}{2}$$ Therefore, $\tan \frac{B+C}{2} = \cot \frac{A}{2}$.