Let α and β be nonzero real numbers such that 2(cosβ - cosα) + cosα cosβ = 1. Then which of the foll

Q: Why do we use the half-angle identity instead of just solving for $\cos\alpha$ or $\cos\beta$ directly?

Great question! While you could solve for the angles, the options are explicitly in terms of $\tan(\alpha/2)$ and $\tan(\beta/2)$. Using the half-angle identity is the most direct path to the target form, saving you from complex inverse trigonometric manipulations. It transforms a transcendental equation into a manageable algebraic one.

Q: I got stuck during the cross-multiplication. Is there a trick to avoid errors?

Algebraic stamina is key here. When you cross-multiply, write out every term clearly. Notice that the constant terms and the product terms will cancel out, leaving you with a clean relationship between the squares of the tangents. If you don't see the cancellation, re-check your signs! The symmetry of the equation is your best friend.

Q: Why do we have both positive and negative roots in the final answer?

Remember that when we have $x^2 = k$, the solution is $x = \pm \sqrt{k}$. In trigonometry, this corresponds to the two possible relationships between the angles that satisfy the original constraint. Both are valid, which is why the question asks for 'which of the following is/are true'. Always respect the square root property!

Question

Let $\alpha$ and $\beta$ be nonzero real numbers such that $2(\cos\beta - \cos\alpha) + \cos\alpha \cos\beta = 1$. Then which of the following is/are true?

Options:

(A) $	an\left(\frac{\alpha}{2}ight) + \sqrt{3}	an\left(\frac{\beta}{2}ight) = 0$

(B) $\sqrt{3}	an\left(\frac{\alpha}{2}ight) + 	an\left(\frac{\beta}{2}ight) = 0$

(C) $	an\left(\frac{\alpha}{2}ight) - \sqrt{3}	an\left(\frac{\beta}{2}ight) = 0$

(D) $\sqrt{3}	an\left(\frac{\alpha}{2}ight) - 	an\left(\frac{\beta}{2}ight) = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow learners. Today, we are going to dismantle a problem that, at first glance, might seem like a chaotic mess of trigonometric functions. You see an equation like $2(\cos\beta - \cos\alpha) + \cos\alpha \cos\beta = 1$, and your instinct might be to panic.

You might think, "How on earth do I relate these two angles, $\alpha$ and $\beta$, to their half-angle tangents?" But I want you to take a deep breath. In JEE Advanced, the complexity of an expression is often a mask for a hidden, beautiful symmetry. Our job is not to fight the equation, but to dance with it.

### The Art of Isolation

Let us look at the given equation: $2(\cos\beta - \cos\alpha) + \cos\alpha \cos\beta = 1$. The first step in any complex problem is to simplify the landscape. Let's expand those brackets:
$$2\cos\beta - 2\cos\alpha + \cos\alpha \cos\beta = 1$$

Now, look at the terms. We have $\cos\beta$ appearing in two places. Let's group them:
$$2\cos\beta + \cos\alpha \cos\beta = 1 + 2\cos\alpha$$

We can factor out $\cos\beta$ on the left side:
$$\cos\beta (2 + \cos\alpha) = 1 + 2\cos\alpha$$

With a simple division, we isolate our variable:
$$\cos\beta = \frac{1 + 2\cos\alpha}{2 + \cos\alpha}$$
This is a massive victory. We have successfully expressed $\cos\beta$ entirely in terms of $\cos\alpha$.

### The Bridge to Half-Angles

Now, we face the next hurdle. The options provided are in terms of $	an(\alpha/2)$ and $	an(\beta/2)$. This is where the Weierstrass substitution, or the half-angle identity, becomes our most powerful tool. Recall the identity:
$$\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$$

Let's define $t_\alpha = 	an(\alpha/2)$ and $t_\beta = 	an(\beta/2)$ to keep our algebra clean. Substituting these into our isolated equation, we get:
$$\frac{1 - t_\beta^2}{1 + t_\beta^2} = \frac{1 + 2\left(\frac{1 - t_\alpha^2}{1 + t_\alpha^2}ight)}{2 + \frac{1 - t_\alpha^2}{1 + t_\alpha^2}}$$

Let's tackle the numerator and denominator of the right-hand side separately. For the numerator, we have:
$$1 + \frac{2(1 - t_\alpha^2)}{1 + t_\alpha^2} = \frac{1 + t_\alpha^2 + 2 - 2t_\alpha^2}{1 + t_\alpha^2} = \frac{3 - t_\alpha^2}{1 + t_\alpha^2}$$

Similarly, the denominator becomes:
$$2 + \frac{1 - t_\alpha^2}{1 + t_\alpha^2} = \frac{2 + 2t_\alpha^2 + 1 - t_\alpha^2}{1 + t_\alpha^2} = \frac{3 + t_\alpha^2}{1 + t_\alpha^2}$$

### The Beauty of Cancellation

Now, watch the magic happen. When we put the numerator and denominator back together, we have:
$$\frac{1 - t_\beta^2}{1 + t_\beta^2} = \frac{\frac{3 - t_\alpha^2}{1 + t_\alpha^2}}{\frac{3 + t_\alpha^2}{1 + t_\alpha^2}}$$

The term $(1 + t_\alpha^2)$ is in the denominator of both the numerator and the denominator of the right-hand side. They cancel out perfectly! We are left with:
$$\frac{1 - t_\beta^2}{1 + t_\beta^2} = \frac{3 - t_\alpha^2}{3 + t_\alpha^2}$$

### The Final Stretch

Cross-multiply to clear the fractions:
$$(1 - t_\beta^2)(3 + t_\alpha^2) = (1 + t_\beta^2)(3 - t_\alpha^2)$$

Expanding both sides, we get:
$$3 + t_\alpha^2 - 3t_\beta^2 - t_\alpha^2 t_\beta^2 = 3 - t_\alpha^2 + 3t_\beta^2 - t_\alpha^2 t_\beta^2$$

Look at the symmetry! The $3$ on both sides cancels, and the $-t_\alpha^2 t_\beta^2$ on both sides cancels. We are left with:
$$t_\alpha^2 - 3t_\beta^2 = -t_\alpha^2 + 3t_\beta^2$$

Grouping the terms gives us $2t_\alpha^2 = 6t_\beta^2$, or $t_\alpha^2 = 3t_\beta^2$. Taking the square root, we get $t_\alpha = \pm \sqrt{3} t_\beta$. This leads us to two possible relations:
$$	an(\alpha/2) - \sqrt{3}	an(\beta/2) = 0 \quad 	ext{or} \quad 	an(\alpha/2) + \sqrt{3}	an(\beta/2) = 0$$

**Final Answer:** The relationship is **$	an(\alpha/2) = \pm \sqrt{3} 	an(\beta/2)$**.

Animated Solution for Mathematics - Trigonometry: Let $α$ and $β$ be nonzero real numbers such that $2 (cos β - cos α) + cos α cos β = 1$ . Then which of the following is/are true?

Select Answer:

Visualized Solution

Grouping Terms

Isolating $cos β$

The Half-Angle Identity

Substituting Variables

Taking Common Denominators

Simplifying Numerators

Canceling Common Terms

Cross Multiplication

Expanding Both Sides

Canceling Identical Terms

Grouping Like Terms

Simplifying the Relation

Taking the Square Root

Matching with Options

The Sigma Insight: Multiple and Sub-multiple Angles

Analyzing the Setup

The Art of Isolation

The Bridge to Half-Angles

The Beauty of Cancellation

The Final Stretch

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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; } Let α and β be nonzero real numbers such that 2(cosβ−cosα)+cosαcosβ=1. Then which of the following is/are true?