In any triangle ABC, prove that cot A/2 + cot B/2 + cot C/2 = cot A/2 cot B/2 cot C/2.

Q: Why do we use $\tan$ instead of $\cot$ in the intermediate steps? Can't we just use $\cot$ directly?

That is a fantastic question! While you *could* use $\cot$ addition formulas, they are often less intuitive and prone to sign errors. The tangent addition formula is a standard, well-behaved tool in our arsenal. We use $\tan$ to simplify the algebra into a clean, manageable form, and then perform a simple conversion to $\cot$ at the very end. It is about choosing the path of least resistance.

Q: Is it valid to divide by $\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}$? What if one of the tangents is zero?

In any triangle, the angles $A, B, C$ are strictly between $0$ and $\pi$. Therefore, the half-angles are strictly between $0$ and $\frac{\pi}{2}$. Since $\tan \theta$ is strictly positive in the first quadrant, the product of tangents is never zero. Division is perfectly safe and mathematically rigorous here!

Q: Why is this identity so important for JEE? Is it just for memorization?

This identity is not just a formula to memorize; it is a gateway to understanding the symmetry of triangle properties. It appears frequently in problems involving the incenter, excircles, and complex geometric configurations. Mastering this manipulation builds the algebraic intuition required for much harder problems involving cyclic sums and symmetric expressions.

Question

In any triangle $ABC$, prove that $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$.

Sigma Percentile · Accepted Answer

### The Geometry of Symmetry: Unlocking the Triangle Identity

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a trigonometric identity; we are uncovering a hidden symmetry that exists within every single triangle in the universe.

We are going to prove that for any triangle $ABC$, the sum of the cotangents of the half-angles is equal to their product:

$$\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$$

### Phase 1: The Foundation

Every great proof begins with a simple truth. For any triangle $ABC$, the sum of the interior angles is constant: $A + B + C = \pi$. This is the DNA of our problem.

However, our target identity involves half-angles. So, let us divide this fundamental equation by $2$ to get:

$$\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}$$

Now, we need to isolate two angles on one side to prepare for a trigonometric operation. Let us move $\frac{C}{2}$ to the right side:

$$\frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2}$$

This simple rearrangement is the key that unlocks the door to the trigonometric world.

### Phase 2: The Trigonometric Bridge

Why do we apply the tangent function here? You might be tempted to use cotangent, but the tangent addition formula is our most reliable ally. Let us take the tangent of both sides:

$$	an\left(\frac{A}{2} + \frac{B}{2}ight) = 	an\left(\frac{\pi}{2} - \frac{C}{2}ight)$$

Here, we invoke the beauty of complementary angle identities. We know that $	an(\frac{\pi}{2} - 	heta) = \cot 	heta$. Therefore, our right side simplifies beautifully to $\cot \frac{C}{2}$.

Our equation now stands as:

$$	an\left(\frac{A}{2} + \frac{B}{2}ight) = \cot \frac{C}{2}$$

### Phase 3: The Algebraic Alchemy

Now, let us expand the left side using the tangent addition formula:

$$\frac{	an \frac{A}{2} + 	an \frac{B}{2}}{1 - 	an \frac{A}{2} 	an \frac{B}{2}} = \cot \frac{C}{2}$$

To make this equation uniform, we convert the right side into its reciprocal form:

$$\frac{	an \frac{A}{2} + 	an \frac{B}{2}}{1 - 	an \frac{A}{2} 	an \frac{B}{2}} = \frac{1}{	an \frac{C}{2}}$$

Now, we cross-multiply. This is where the magic happens. We get:

$$	an \frac{C}{2} \left(	an \frac{A}{2} + 	an \frac{B}{2}ight) = 1 - 	an \frac{A}{2} 	an \frac{B}{2}$$

Distributing the $	an \frac{C}{2}$ gives us:

$$	an \frac{A}{2} 	an \frac{C}{2} + 	an \frac{B}{2} 	an \frac{C}{2} = 1 - 	an \frac{A}{2} 	an \frac{B}{2}$$

Rearranging the terms, we arrive at a standard, incredibly useful identity:

$$	an \frac{A}{2} 	an \frac{B}{2} + 	an \frac{B}{2} 	an \frac{C}{2} + 	an \frac{C}{2} 	an \frac{A}{2} = 1$$

Memorize this! It is a cornerstone of triangle trigonometry.

### Phase 4: The Final Transformation

We are almost there. We have an identity in tangents, but we need cotangents. We divide every single term on both sides by the product $	an \frac{A}{2} 	an \frac{B}{2} 	an \frac{C}{2}$.

Watch the cancellation:

$$\frac{	an \frac{A}{2} 	an \frac{B}{2}}{	an \frac{A}{2} 	an \frac{B}{2} 	an \frac{C}{2}} + \frac{	an \frac{B}{2} 	an \frac{C}{2}}{	an \frac{A}{2} 	an \frac{B}{2} 	an \frac{C}{2}} + \frac{	an \frac{C}{2} 	an \frac{A}{2}}{	an \frac{A}{2} 	an \frac{B}{2} 	an \frac{C}{2}} = \frac{1}{	an \frac{A}{2} 	an \frac{B}{2} 	an \frac{C}{2}}$$

Each term simplifies perfectly into a cotangent. The first term becomes $\frac{1}{	an \frac{C}{2}} = \cot \frac{C}{2}$. The second becomes $\cot \frac{A}{2}$, and the third becomes $\cot \frac{B}{2}$.

The right side becomes the product of the cotangents. Thus, we arrive at our destination:

$$\mathbf{\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} = \cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}}$$

Take a moment to breathe and appreciate the elegance of this result. We started with a simple triangle and, through logical steps, revealed a profound symmetry. This is the essence of JEE mathematics—not just calculation, but the discovery of order in complexity.

Animated Solution for Mathematics - Trigonometry: In any triangle $A B C$ , prove that $cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$ .

Visualized Solution

$A + B + C = π$

$\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{π}{2}$

$\frac{A}{2} + \frac{B}{2} = \frac{π}{2} - \frac{C}{2}$

$tan (\frac{A}{2} + \frac{B}{2})$

$tan (\frac{π}{2} - θ) = cot θ$

$tan (x + y)$ Expansion

$cot θ = \frac{1}{t a n θ}$

Cross-Multiplication

Distributing Terms

$\sum tan \frac{A}{2} tan \frac{B}{2} = 1$

Dividing by Product

Simplifying Fractions

Final Identity

The Sigma Insight: Conditional Identities

The Geometry of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In any triangle ABC, prove that cot2A​+cot2B​+cot2C​=cot2A​cot2B​cot2C​.