Let θ, ϕ ∈ [0, 2π] be such that 2 cos θ (1 - sin ϕ) = sin² θ (tan θ/2 + cot θ/2) cos ϕ - 1, tan(2π -

Q: Wait, why do we reject the second quadrant for $\theta$ when $\tan(2\pi - \theta) > 0$?

That is a fantastic observation! Let's break it down. The condition $\tan(2\pi - \theta) > 0$ simplifies to $-\tan \theta > 0$, which means $\tan \theta < 0$. Tangent is negative in the second and fourth quadrants. However, we have a second constraint: $-1 < \sin \theta < -\sqrt{3}/2$. Sine is negative in the third and fourth quadrants. The only quadrant that satisfies both conditions—where tangent is negative AND sine is negative—is the fourth quadrant. Always look for the intersection of all constraints!

Q: Why did we choose $n=1$ for the general solution of $\sin(\theta + \phi)$?

We are looking for the range of $\theta + \phi$. We know $\theta$ is in the fourth quadrant, specifically between $3\pi/2$ and $5\pi/3$. Since $\phi$ is also in $[0, 2\pi]$, their sum $\theta + \phi$ will naturally push into the second rotation of the unit circle. The general solution for $\sin x \in (1/2, 1)$ is $(2n\pi + \pi/6, 2n\pi + 5\pi/6)$. For $n=0$, the values are too small. For $n=1$, we get $(13\pi/6, 17\pi/6)$, which perfectly aligns with the range of $\theta + \phi$ given our constraints on $\theta$.

Q: How do we handle the subtraction of intervals for $\phi$?

When you have an inequality like $\phi = (\theta + \phi) - \theta$, you are essentially finding the bounds of a difference. To find the minimum $\phi$, you take the minimum of $(\theta + \phi)$ and subtract the maximum of $\theta$. To find the maximum $\phi$, you take the maximum of $(\theta + \phi)$ and subtract the minimum of $\theta$. It is a simple logic: to make a difference as small as possible, subtract the largest possible value from the smallest possible value!

Question

Let $	heta, \phi \in [0, 2\pi]$ be such that $2 \cos 	heta (1 - \sin \phi) = \sin^2 	heta (	an \frac{	heta}{2} + \cot \frac{	heta}{2}) \cos \phi - 1$, $	an(2\pi - 	heta) > 0$ and $-1 < \sin 	heta < -\sqrt{3}/2$, then $\phi$ cannot satisfy

Options:

(A) $0 < \phi < \pi/2$

(B) $\pi/2 < \phi < 4\pi/3$

(C) $4\pi/3 < \phi < 3\pi/2$

(D) $3\pi/2 < \phi < 2\pi$

Sigma Percentile · Accepted Answer

### The Art of Trigonometric Reduction Welcome, warriors of JEE Advanced! Today, we are not just solving a trigonometric equation; we are peeling back the layers of a mathematical onion. When you first look at an equation like $$2 \cos heta (1 - \sin \phi) = \sin^2 heta \left( an \frac{ heta}{2} + \cot \frac{ heta}{2} ight) \cos \phi - 1$$ it is natural to feel a surge of intimidation. But remember, in the world of JEE, complexity is often just a mask for elegance. Let us strip away that mask together. ### Phase 1: Taming the Beast The most terrifying part of this equation is the term $ an \frac{ heta}{2} + \cot \frac{ heta}{2}$. Let us apply the fundamental definitions where $ an x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$. When we add these, we get: $$\frac{\sin^2( heta/2) + \cos^2( heta/2)}{\sin( heta/2)\cos( heta/2)}$$ The numerator is the beautiful Pythagorean identity, which equals $1$. The denominator is $\sin( heta/2)\cos( heta/2)$, which, by the double angle formula $\sin 2A = 2 \sin A \cos A$, is simply $\frac{1}{2} \sin heta$. Thus, our entire complex term simplifies to $\frac{2}{\sin heta}$. Suddenly, the equation is breathing again! ### Phase 2: The Transformation Now, let us substitute this back into our main equation. The $\sin^2 heta$ term on the right side meets our new $\frac{2}{\sin heta}$, and they cancel out perfectly, leaving us with $2 \sin heta \cos \phi - 1$. Our equation now reads: $$2 \cos heta (1 - \sin \phi) = 2 \sin heta \cos \phi - 1$$ If we expand the left side, we get $2 \cos heta - 2 \cos heta \sin \phi = 2 \sin heta \cos \phi - 1$. Rearranging this, we bring the terms involving $\phi$ to one side: $$2 \cos heta + 1 = 2 \sin heta \cos \phi + 2 \cos heta \sin \phi$$ Do you see it? The right side is the classic expansion of $\sin(A + B)$. We have reduced the entire expression to: $$2 \cos heta + 1 = 2 \sin( heta + \phi)$$ This is the heart of the problem. ### Phase 3: The Constraint Trap Now, we must respect the boundaries. We are given $ an(2\pi - heta) > 0$ and $-1 < \sin heta < -\frac{\sqrt{3}}{2}$. The first condition implies $- an heta > 0$, so $ an heta < 0$. This restricts $ heta$ to the second or fourth quadrant. The second condition, $\sin heta < 0$, restricts $ heta$ to the third or fourth quadrant. The intersection is the fourth quadrant! Specifically, since $\sin heta$ ranges from $-1$ to $-\frac{\sqrt{3}}{2}$, $ heta$ must lie in the interval $(\frac{3\pi}{2}, \frac{5\pi}{3})$. This is the geometric reality of our variable $ heta$. ### Phase 4: The Final Range With $ heta \in (\frac{3\pi}{2}, \frac{5\pi}{3})$, we can find the range of $2 \cos heta + 1$. In this interval, $\cos heta$ ranges from $0$ to $\frac{1}{2}$. Therefore, $2 \cos heta$ ranges from $0$ to $1$, and $2 \cos heta + 1$ ranges from $1$ to $2$. Since $2 \sin( heta + \phi) = 2 \cos heta + 1$, we have: $$1 < 2 \sin( heta + \phi) < 2 \implies \frac{1}{2} < \sin( heta + \phi) < 1$$ Using the general solution for sine, we find $ heta + \phi \in (\frac{13\pi}{6}, \frac{17\pi}{6})$. Finally, subtracting the range of $ heta$ from this sum gives us the range of $\phi$: **$$\phi \in \left( \frac{\pi}{2}, \frac{4\pi}{3} ight)$$** Any value outside this range cannot satisfy the equation. You have conquered the problem!

Animated Solution for Mathematics - Trigonometry: Let $θ, ϕ \in [0, 2 π]$ be such that $2 cos θ (1 - sin ϕ) = sin^{2} θ (tan \frac{θ}{2} + cot \frac{θ}{2}) cos ϕ - 1$ , $tan (2 π - θ) > 0$ and $- 1 < sin θ < - 3 /2$ , then $ϕ$ cannot satisfy

Select Answer:

Visualized Solution

Visualizing the Problem

Simplifying the Identity $tan \frac{θ}{2} + cot \frac{θ}{2}$

Substituting Back into the Equation

Expanding and Rearranging

Applying the Compound Angle Formula

Analyzing Constraints on $θ$

Determining the Range of $θ$

Finding the Range of $2 cos θ + 1$

Solving for $sin (θ + ϕ)$

Determining the Range of $θ + ϕ$

Calculating the Final Range of $ϕ$

Conclusion and Excluded Intervals

The Sigma Insight: General Solution of Trigonometric Equations

The Art of Trigonometric Reduction

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Visualizing the Problem

Simplifying the Identity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan2θ​+cot2θ​

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