Let α and β are two real roots of the equation (k + 1) tan² x - √2λ tan x = 1 - k, where (k ≠ -1) an

Q: Wait, why do we use $\tan \alpha$ and $\tan \beta$ as the roots of the quadratic? Shouldn't the roots be $\alpha$ and $\beta$?

That is a fantastic question! The equation is given in terms of $\tan x$. If $x = \alpha$ is a root of the equation, then $\tan \alpha$ must satisfy the equation. Therefore, $\tan \alpha$ and $\tan \beta$ are the values that satisfy the quadratic equation in $u = \tan x$. We are solving for the roots of the quadratic in $u$, which are the tangents of the original roots.

Q: Why can't we just solve for $x$ directly?

We cannot solve for $x$ directly because we have two unknown parameters, $k$ and $\lambda$. We don't need the individual values of $\alpha$ and $\beta$; we only need their relationship through the tangent addition formula. This is the beauty of JEE problems—they often test your ability to manipulate relationships rather than finding individual variables.

Q: Does the condition $k \neq -1$ actually matter?

Yes, it is crucial! If $k = -1$, the coefficient of $\tan^2 x$ becomes zero, and the equation is no longer a quadratic. It would collapse into a linear equation, and our entire strategy of using Vieta's formulas would fail. Always check the constraints provided in the problem statement!

Question

Let $\alpha$ and $\beta$ are two real roots of the equation $(k + 1) 	an^2 x - \sqrt{2}\lambda 	an x = 1 - k$, where $(k 
eq -1)$ and $\lambda$ are real numbers. If $	an^2(\alpha + \beta) = 50$, then value of $\lambda$ is

Options:

(A) $5\sqrt{2}$

(B) $10\sqrt{2}$

(C) $10$

(D) $5$

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Trigonometric Quadratics

Welcome, future engineer. Today, we are going to peel back the layers of a problem that, at first glance, looks like a chaotic mess of trigonometry and algebra.

But as you will soon see, beneath the surface lies a beautiful, elegant structure waiting to be revealed. Let us embark on this journey together.

### Phase 1: Unmasking the Quadratic

Imagine you are standing before the equation: 
$$(k + 1) 	an^2 x - \sqrt{2}\lambda 	an x = 1 - k$$

It looks intimidating, doesn't it? We have $	an^2 x$ and $	an x$ mixed with parameters $k$ and $\lambda$.

The secret to mastering such problems is to stop seeing the trigonometry for a moment. If we let $u = 	an x$, the equation transforms into:
$$(k + 1)u^2 - \sqrt{2}\lambda u + (k - 1) = 0$$

Suddenly, the fog clears. You are not looking at a complex trigonometric equation; you are looking at a standard quadratic equation of the form $au^2 + bu + c = 0$. This is the first step in any JEE problem: identify the structure.

### Phase 2: The Power of Vieta

Now that we have our quadratic, we know that $\alpha$ and $\beta$ are the roots of the original equation. This means that $u_1 = 	an \alpha$ and $u_2 = 	an \beta$ are the roots of our quadratic in $u$.

Here is where we pull out our most powerful tool: **Vieta's Formulas**. We know that for any quadratic $au^2 + bu + c = 0$, the sum of the roots is $-b/a$ and the product is $c/a$.

Applying this to our equation, we get the sum:
$$	an \alpha + 	an \beta = \frac{\sqrt{2}\lambda}{k + 1}$$

And the product:
$$	an \alpha 	an \beta = \frac{k - 1}{k + 1}$$

These two expressions are the keys to the kingdom. Hold onto them tightly.

### Phase 3: The Trigonometric Bridge

The problem gives us a condition: $	an^2(\alpha + \beta) = 50$. This is our target.

How do we connect our sum and product to $	an(\alpha + \beta)$? We use the compound angle identity:
$$	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$$

This identity is the bridge that connects the algebraic roots to the trigonometric condition. Let us substitute our Vieta expressions into this formula.

The numerator becomes $\frac{\sqrt{2}\lambda}{k + 1}$, and the denominator becomes $1 - \frac{k - 1}{k + 1}$.

### Phase 4: The Elegant Cancellation

Now, let us simplify the denominator. It looks messy, but watch what happens:
$$1 - \frac{k - 1}{k + 1} = \frac{(k + 1) - (k - 1)}{k + 1} = \frac{2}{k + 1}$$

When we divide the numerator by this denominator, the $(k + 1)$ terms in the denominators cancel out perfectly!

We are left with:
$$	an(\alpha + \beta) = \frac{\sqrt{2}\lambda}{2} = \frac{\lambda}{\sqrt{2}}$$

Is that not beautiful? All that complexity collapsed into a simple expression involving only $\lambda$.

### The Final Victory

We are almost there. We are given that $	an^2(\alpha + \beta) = 50$.

Substituting our result, we get:
$$\left( \frac{\lambda}{\sqrt{2}} ight)^2 = 50$$

This simplifies to $\frac{\lambda^2}{2} = 50$. This leads us directly to $\lambda^2 = 100$, and since we are looking for the positive value, **$\lambda = 10$**.

You have successfully navigated the trap, used the right tools, and arrived at the solution. Remember, in JEE, it is rarely about brute force; it is about finding the elegant path through the complexity.

Animated Solution for Mathematics - Trigonometry: Let $α$ and $β$ are two real roots of the equation $(k + 1) tan^{2} x - 2 λ tan x = 1 - k$ , where $(k \neq = - 1)$ and $λ$ are real numbers. If $tan^{2} (α + β) = 50$ , then value of $λ$ is

Select Answer:

Visualized Solution

Standard Quadratic Form

Roots of the Quadratic

Sum of Roots

Product of Roots

Compound Angle Identity

Substitute Sum and Product

Simplify the Denominator

Simplified Expression

Apply Given Condition

Solve for $λ$

The Sigma Insight: Trigonometric Ratios and Identities

The Hidden Symmetry of Trigonometric Quadratics

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

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