If 0

Q: Why do we square both sides? Doesn't that introduce extraneous roots?

That is a brilliant observation! Yes, squaring can introduce extraneous roots. However, in this context, it is the most efficient way to convert the sum $\cos x + \sin x$ into a form involving $\sin 2x$. We simply need to verify our final values against the original equation $\cos x + \sin x = 1/2$ to ensure they are valid.

Q: Why is $\sin 2x$ negative? I thought $x$ was between $0$ and $\pi$.

Great catch. If $x \in (0, \pi)$, then $2x \in (0, 2\pi)$. Our calculation showed $\sin 2x = -3/4$. Since the sine is negative, $2x$ must lie in the third or fourth quadrant (i.e., between $\pi$ and $2\pi$). This forces $x$ to be in the second quadrant, specifically between $\pi/2$ and $\pi$.

Q: How do I know which of the two roots for $\tan x$ to pick?

We use the constraint $\cos x + \sin x = 1/2$. In the second quadrant, $\sin x$ is positive and $\cos x$ is negative. For their sum to be positive, $\sin x$ must be greater than the magnitude of $\cos x$. This implies $\tan x$ must be less than $-1$. By testing our two roots, we can easily see which one satisfies this condition.

Question

If $0 < x < \pi$ and $\cos x + \sin x = 1/2$, then $	an x$ is

Options:

(A) $\frac{(1-\sqrt{7})}{4}$

(B) $\frac{(4-\sqrt{7})}{3}$

(C) $-\frac{(4+\sqrt{7})}{3}$

(D) $\frac{(1+\sqrt{7})}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are not just solving an equation; we are embarking on a geometric journey. We are given the elegant constraint $0 < x < \pi$ and the intriguing equation $\cos x + \sin x = 1/2$.

At first glance, this looks like a simple algebraic puzzle, but beneath the surface lies a beautiful interplay between the unit circle and the tangent function. Let us peel back the layers together.

### The Squaring Intuition

When you see a sum of sine and cosine, your mathematical intuition should immediately scream, "Square me!" We know that $(\cos x + \sin x)^2 = \cos^2 x + \sin^2 x + 2 \sin x \cos x$.

This is the golden key. The term $\cos^2 x + \sin^2 x$ collapses into the number $1$, and $2 \sin x \cos x$ is the famous double angle identity, $\sin 2x$.

By squaring both sides of $\cos x + \sin x = 1/2$, we get $(\cos x + \sin x)^2 = (1/2)^2$, which expands to $1 + \sin 2x = 1/4$. Suddenly, the complexity vanishes, and we are left with:

$$\sin 2x = -3/4$$

This is our first major milestone.

### The Double Angle Bridge

Now, the question asks for $	an x$. We have $\sin 2x$, but we need $	an x$. How do we bridge this gap?

We invoke the powerful identity:

$$\sin 2x = \frac{2 	an x}{1 + 	an^2 x}$$

This identity is the secret passage that connects the world of double angles to the world of tangents. Let us set $t = 	an x$ to keep our algebra clean. Substituting our value, we get:

$$\frac{2t}{1 + t^2} = -3/4$$

Cross-multiplying gives us $8t = -3(1 + t^2)$, which rearranges into the quadratic equation:

$$3t^2 + 8t + 3 = 0$$

Take a deep breath. We have arrived at the heart of the problem.

### The Quadratic Trap and the Final Selection

Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find:

$$t = \frac{-8 \pm \sqrt{64 - 36}}{6} = \frac{-8 \pm \sqrt{28}}{6} = \frac{-4 \pm \sqrt{7}}{3}$$

We have two candidates for $	an x$. But which one is the true solution? This is where the JEE tests your depth of understanding.

We established that $\sin 2x = -3/4$, which is negative. This places $2x$ in the third or fourth quadrant, meaning $x$ must be in the second quadrant, where $\sin x > 0$ and $\cos x < 0$.

Furthermore, because $\cos x + \sin x = 1/2$ (a positive value), the positive sine component must be larger than the magnitude of the negative cosine component. This forces $	an x < -1$.

Evaluating our roots, $\frac{-4 + \sqrt{7}}{3} \approx -0.45$ and $\frac{-4 - \sqrt{7}}{3} \approx -2.21$. Only the second value is less than $-1$.

Thus, we have our final answer:

$$	an x = \frac{-(4 + \sqrt{7})}{3}$$

You have navigated the geometry, the identities, and the quadrant constraints with grace. This is the essence of JEE Advanced mathematics—not just calculation, but careful, logical navigation.

Animated Solution for Mathematics - Trigonometry: If $0 < x < π$ and $cos x + sin x = 1/2$ , then $tan x$ is

Select Answer:

Visualized Solution

Visualizing the Constraint

Squaring the Equation

Expanding the Square

Applying Trigonometric Identities

Isolating $sin 2 x$

Expressing in terms of $tan x$

Forming a Quadratic Equation

Applying the Quadratic Formula

Simplifying the Roots

Analyzing the Quadrant

Refining the Interval

Selecting the Correct Value

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Squaring Intuition

The Double Angle Bridge

The Quadratic Trap and the Final Selection

Similar Questions

If $\frac{2 s i n α}{1 + c o s 2 α} = \frac{1}{7}$ and $\frac{1 - c o s 2 β}{2} = \frac{1}{10}$ , $α, β \in (0, \frac{π}{2})$ , then $tan (α + 2 β)$ is equal to _________ .

If $\frac{s i n ^{4} x}{2} + \frac{c o s ^{4} x}{3} = \frac{1}{5}$ , then

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

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

If $cot x = \frac{5}{12}$ for some $x \in (π, \frac{3 π}{2})$ , then $sin 7 x (cos \frac{13 x}{2} + sin \frac{13 x}{2}) + cos 7 x (cos \frac{13 x}{2} - sin \frac{13 x}{2})$ is equal to

For $α, β \in (0, \frac{π}{2})$ , let $3 sin (α + β) = 2 sin (α - β)$ and a real number $k$ be such that $tan α = k tan β$ . Then the value of $k$ is equal to :

If $tan θ = - 4/3$ , then $sin θ$ is

If $sin x = - \frac{3}{5}$ , where $π < x < \frac{3 π}{2}$ , then $80 (tan^{2} x - cos x)$ is equal to

If $s in x + s i n^{2} x = 1, x \in (0, \frac{π}{2})$ , then $(co s^{12} x + t a n^{12} x) + 3 (co s^{10} x + t a n^{10} x + co s^{8} x + t a n^{8} x) + (co s^{6} x + t a n^{6} x)$ is equal to:

In a $△ P QR$ , if $3 sin P + 4 cos Q = 6$ and $4 sin Q + 3 cos P = 1$ , then the angle $R$ is equal to

Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 0<x<π and cosx+sinx=1/2, then tanx is