sec² θ = 4xy / (x+y)² is true if and only if (A) x+y ≠ 0 (B) x=y, x ≠ 0 (C) x=y (D) x ≠ 0, y ≠ 0.

Q: Wait, why do we reject the negative root here? Or rather, why is $(x-y)^2 \le 0$ only possible when $x=y$?

That is a fantastic question! Remember that for any real number $a$, the square $a^2$ is always greater than or equal to zero. It can never be negative. So, if we have an inequality stating that a square is less than or equal to zero, the only way it can be true is if that square is exactly zero. This forces the term inside the square to be zero, leading us directly to $x=y$.

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm! However, this is a pure mathematical identity problem. In JEE Advanced, you will often encounter problems that look like physics equations but are actually testing your grasp of algebraic constraints and trigonometric ranges. Always look at the mathematical structure first—if there is no time or motion involved, it is likely a test of your algebraic intuition.

Q: Why do we need $x \neq 0$ at the end? Is it just a formality?

It is absolutely not a formality! In mathematics, especially in JEE, the domain of an expression is just as important as the equation itself. The original expression $\frac{4xy}{(x+y)^2}$ is undefined if the denominator is zero. Since we found $x=y$, the denominator becomes $(x+x)^2 = 4x^2$. If $x$ were zero, the denominator would be zero, making the expression undefined. Thus, $x \neq 0$ is a crucial constraint.

Question

$\sec^2 	heta = \frac{4xy}{(x+y)^2}$ is true if and only if

Options:

(A) $x+y 
eq 0$

(B) $x=y, x 
eq 0$

(C) $x=y$

(D) $x 
eq 0, y 
eq 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex equation: $\sec^2 	heta = \frac{4xy}{(x+y)^2}$. At first glance, it looks like a standard trigonometric problem.

But as an elite JEE aspirant, you know that appearances can be deceiving. This is not just about finding an angle $	heta$; it is about uncovering the rigid algebraic structure hidden beneath the surface of trigonometry.

### The Range Trap

The first step in any problem involving trigonometric functions equated to algebraic expressions is to pause and consider the range. We know that for any real angle $	heta$, the cosine function is bounded: $-1 \le \cos 	heta \le 1$.

Consequently, the secant function, being the reciprocal, must satisfy $|\sec 	heta| \ge 1$. When we square this, we get the fundamental truth: $\sec^2 	heta \ge 1$.

This is our anchor. If the left side of our equation is always at least one, then the right side must also be at least one. This simple realization transforms our trigonometric problem into a powerful algebraic inequality:

$$\frac{4xy}{(x+y)^2} \ge 1$$

### The Algebraic Bridge

Now, we must handle this inequality with care. We know that $(x+y)^2$ is a perfect square, which means it is always non-negative. As long as $x+y 
eq 0$, it is strictly positive.

Multiplying both sides by $(x+y)^2$ is safe, and the inequality sign remains unchanged:

$$4xy \ge (x+y)^2$$

Now, let us expand the right side using the identity $(x+y)^2 = x^2 + 2xy + y^2$. Our inequality becomes:

$$4xy \ge x^2 + 2xy + y^2$$

### The Perfect Square Revelation

Let us bring all terms to one side to see the structure more clearly. Subtracting $4xy$ from both sides, we get:

$$0 \ge x^2 + 2xy + y^2 - 4xy$$

This simplifies to:

$$0 \ge x^2 - 2xy + y^2$$

Look closely at that expression on the right. It is a perfect square! It condenses beautifully into $(x-y)^2$. So, we are left with the condition:

$$(x-y)^2 \le 0$$

We know that the square of any real number cannot be negative. Therefore, the only way for $(x-y)^2$ to be less than or equal to zero is if it is exactly equal to zero. This implies $x - y = 0$, or simply **$x = y$**.

### The Domain Guard

We have found that $x=y$, but we must be vigilant. We must check the original equation for any hidden constraints. The denominator $(x+y)^2$ cannot be zero.

If we substitute $x=y$ into the denominator, we get $(x+x)^2 = (2x)^2 = 4x^2$. For this to be non-zero, $x$ cannot be zero.

This is the final piece of the puzzle. The equation holds true if and only if **$x = y$ and $x 
eq 0$**.

Animated Solution for Mathematics - Trigonometry: $sec^{2} θ = \frac{4 x y}{( x + y ) ^{2}}$ is true if and only if

Select Answer:

Visualized Solution

Analyzing the Equation

Range of $sec^{2} θ$

Setting up the Inequality

Clearing the Denominator

Expanding the Square

Rearranging Terms

Forming a Perfect Square

Analyzing the Perfect Square

Deducing the Relationship

Checking Domain Constraints

Final Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Range Trap

The Algebraic Bridge

The Perfect Square Revelation

The Domain Guard

Similar Questions

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Let $f (θ) = sin θ (sin θ + sin 3 θ)$ . Then $f (θ)$ is

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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; } sec2θ=(x+y)24xy​ is true if and only if

Select Answer:

Visualized Solution

Analyzing the Equation

Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sec2θ

Setting up the Inequality

Clearing the Denominator

Expanding the Square

Rearranging Terms

Forming a Perfect Square

Analyzing the Perfect Square

Deducing the Relationship

Checking Domain Constraints

Final Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Range Trap

The Algebraic Bridge

The Perfect Square Revelation

The Domain Guard

Similar Questions

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Animated Solution for Mathematics - Trigonometry: $sec^{2} θ = \frac{4 x y}{( x + y ) ^{2}}$ is true if and only if

Range of $sec^{2} θ$

If $x = \sum_{n = 0}^{\infty} (- 1)^{n} tan^{2 n} θ$ and $y = \sum_{n = 0}^{\infty} cos^{2 n} θ$ , where $0 < θ < \frac{π}{4}$ , then:

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If $0 < x, y < π$ and $cos x + cos y - cos (x + y) = \frac{3}{2}$ , then $sin x + cos y$ is equal to:

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Given $A = sin^{2} θ + cos^{4} θ$ then for all real values of $θ$

If for $θ \in [- \frac{π}{3}, 0]$ , the points $(x, y) = (3 tan (θ + \frac{π}{3}), 2 tan (θ + \frac{π}{6}))$ lie on $x y + α x + β y + γ = 0$ , then $α^{2} + β^{2} + γ^{2}$ is equal to :

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If $α + β = π /2$ and $β + γ = α$ , then $tan α$ equals

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