If tan A and tan B are the roots of the quadratic equation, 3x²–10x–25=0, then the value of 3 sin²(A

Q: Wait, why can't we just solve for A and B individually using the quadratic formula?

That is a very natural instinct! However, in JEE problems, we often don't need the individual values of A and B. Solving for them might lead to complex or irrational angles that are difficult to work with. By using Vieta's formulas, we bypass the need to find A and B separately, focusing instead on the symmetric functions of the roots, which is much more efficient and elegant.

Q: Why do we divide the expression by $\cos^2(A+B)$? Is that a standard trick?

Yes, it is a classic technique! Whenever you see a homogeneous trigonometric expression—where every term has the same total degree (in this case, degree two)—dividing by the highest power of cosine is the standard 'bridge' to convert the expression into a polynomial in terms of tangent. It turns a complex trigonometric problem into a simple algebraic one.

Q: Is there a faster way to see the answer without all this calculation?

You have a sharp eye! If you look closely at the target expression $3 \sin^2 \theta - 10 \sin \theta \cos \theta - 25 \cos^2 \theta$, it is structurally identical to the quadratic $3x^2 - 10x - 25$ where $x = \tan \theta$. If you divide the expression by $\cos^2 \theta$, you get $\cos^2 \theta (3 \tan^2 \theta - 10 \tan \theta - 25)$. Since $\tan \theta$ is a root of $3x^2 - 10x - 25 = 0$, the bracketed term is actually zero! Wait, let me re-check that... actually, the roots are $\tan A$ and $\tan B$, not $\tan(A+B)$. So, the bracketed term is $3 \tan^2(A+B) - 10 \tan(A+B) - 25$. Since $\tan(A+B)$ is not necessarily a root of the original quadratic, you must perform the calculation. Your intuition is great, but always verify the variable!

Question

If $	an A$ and $	an B$ are the roots of the quadratic equation, $3x^2 - 10x - 25 = 0$, then the value of $3 \sin^2(A+B) - 10 \sin(A+B) \cos(A+B) - 25 \cos^2(A+B)$ is :

Options:

(A) 10

(B) -10

(C) 25

(D) -25

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we are uncovering a hidden symmetry between the world of algebra and the world of trigonometry.

We are given the quadratic equation $3x^2 - 10x - 25 = 0$, with roots $	an A$ and $	an B$. In the high-stakes arena of JEE Advanced, we do not need the individual angles; we need the relationship between them.

### The Power of Vieta

When you see a quadratic equation with roots, your mind should immediately jump to Vieta's formulas. For any quadratic $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product is $\frac{c}{a}$.

Applying this to our equation, we find:

$$	an A + 	an B = -\frac{-10}{3} = \frac{10}{3}$$

$$	an A \cdot 	an B = \frac{-25}{3}$$

We have captured the essence of $A$ and $B$ without ever needing to know their specific values. This is the beauty of mathematical abstraction.

### The Compound Angle Connection

Now, we turn our attention to the target expression:

$$E = 3 \sin^2(A+B) - 10 \sin(A+B) \cos(A+B) - 25 \cos^2(A+B)$$

Notice the angle $(A+B)$ appearing everywhere. We need to evaluate $	an(A+B)$ using the compound angle formula:

$$	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$$

Substituting our Vieta values:

$$	an(A+B) = \frac{10/3}{1 - (-25/3)} = \frac{10/3}{1 + 25/3} = \frac{10/3}{28/3} = \frac{10}{28} = \frac{5}{14}$$

We have successfully bridged the gap. We now know the tangent of the combined angle is $\frac{5}{14}$.

### The Homogeneous Transformation

Notice that every term in $E$ has a degree of two. The 'magic trick' here is to divide by $\cos^2(A+B)$, factoring it out to get:

$$E = \cos^2(A+B) [3 	an^2(A+B) - 10 	an(A+B) - 25]$$

Suddenly, the trigonometric expression has collapsed into a simple quadratic form in terms of $	an(A+B)$. We know $\cos^2(A+B) = \frac{1}{1 + 	an^2(A+B)}$.

Let's calculate the value inside the bracket:

$$3\left(\frac{5}{14}ight)^2 - 10\left(\frac{5}{14}ight) - 25 = 3\left(\frac{25}{196}ight) - \frac{50}{14} - 25$$

$$= \frac{75}{196} - \frac{700}{196} - \frac{4900}{196} = -\frac{5525}{196}$$

Now, for the $\cos^2(A+B)$ term:

$$\cos^2(A+B) = \frac{1}{1 + (5/14)^2} = \frac{1}{1 + 25/196} = \frac{196}{221}$$

### The Grand Finale

We are at the finish line. We multiply our two results:

$$E = \frac{196}{221} 	imes \left(-\frac{5525}{196}ight)$$

The $196$ terms cancel out with satisfying precision, leaving us with:

$$E = -\frac{5525}{221} = -25$$

The final answer is **-25**. This is the thrill of JEE mathematics—the realization that beneath the complexity lies a deep, orderly structure waiting to be revealed by your logic.

Animated Solution for Mathematics - Trigonometry: If $tan A$ and $tan B$ are the roots of the quadratic equation, $3 x^{2} - 10 x - 25 = 0$ , then the value of $3 sin^{2} (A + B) - 10 sin (A + B) cos (A + B) - 25 cos^{2} (A + B)$ is :

Select Answer:

Visualized Solution

Identifying the Quadratic Equation

Applying Vieta's Formulas

Calculating the Sum of Roots

Calculating the Product of Roots

The Compound Angle Formula

Evaluating $tan (A + B)$

Analyzing the Target Expression

Factoring out $cos^{2} (A + B)$

Identity for $cos^{2} θ$

Substituting the Value of $tan (A + B)$

Evaluating the Bracketed Term

Final Calculation and Result

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Power of Vieta

The Compound Angle Connection

The Homogeneous Transformation

The Grand Finale

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Select Answer:

Visualized Solution

Identifying the Quadratic Equation

Applying Vieta's Formulas

Calculating the Sum of Roots

Calculating the Product of Roots

The Compound Angle Formula

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(A+B)

Analyzing the Target Expression

Factoring out .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2(A+B)

Identity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos2θ

Substituting the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(A+B)

Evaluating the Bracketed Term

Final Calculation and Result

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Power of Vieta

The Compound Angle Connection

The Homogeneous Transformation

The Grand Finale

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