cosec 18° is a root of the equation :

Q: Wait, why do we need to rationalize the denominator? Can't we just square $x = \frac{4}{\sqrt{5} - 1}$ directly?

You absolutely could, but it would make the algebra significantly more complex and prone to errors. By rationalizing first, you simplify the expression to $x = \sqrt{5} + 1$. This makes the subsequent step of isolating the radical and squaring it much cleaner and more intuitive.

Q: Is there a faster way to find the value of $\sin 18^\circ$ if I forget it during the exam?

Yes! You can derive it using the identity $\sin 2\theta = \cos 3\theta$ (or $2\theta = 90^\circ - 3\theta$). If $\theta = 18^\circ$, then $2\theta = 36^\circ$ and $3\theta = 54^\circ$. Since $\sin 36^\circ = \cos 54^\circ$, you can solve the resulting quadratic equation for $\sin 18^\circ$. It is a great exercise to practice!

Q: Why do we isolate the radical before squaring? Why not just square $x = \sqrt{5} + 1$ directly?

If you square $x = \sqrt{5} + 1$ directly, you get $x^2 = 5 + 1 + 2\sqrt{5} = 6 + 2\sqrt{5}$. You still have a radical term ($2\sqrt{5}$) on the right side. By isolating the radical first ($x - 1 = \sqrt{5}$), you ensure that when you square both sides, the radical disappears completely, leaving you with a pure polynomial equation.

Question

$	ext{cosec } 18^\circ$ is a root of the equation :

Options:

(A) $x^2 + 2x - 4 = 0$

(B) $4x^2 + 2x - 1 = 0$

(C) $x^2 - 2x + 4 = 0$

(D) $x^2 - 2x - 4 = 0$

Sigma Percentile · Accepted Answer

### The Elegant Dance of Trigonometry and Algebra

Welcome, future engineers! Today, we are going to unravel a beautiful problem that sits at the intersection of two fundamental pillars of mathematics: trigonometry and algebra.

Often, students see a question involving $	ext{cosec } 18^\circ$ and feel a sense of dread, but I want you to see it differently. This is not just a calculation; it is a puzzle waiting to be solved with elegance.

### The Trigonometric Foundation

Our journey begins with the objective: we need to find a quadratic equation where $x = 	ext{cosec } 18^\circ$ is a root.

The first step is to demystify that angle. You should have the value of $\sin 18^\circ$ etched into your memory for the JEE:

$$ \sin 18^\circ = \frac{\sqrt{5} - 1}{4} $$

Think of this as the key to the lock. In a right-angled triangle, this tells us that the ratio of the opposite side to the hypotenuse is $(\sqrt{5} - 1) : 4$.

Since $	ext{cosec } 	heta$ is simply the reciprocal of $\sin 	heta$, we can write our root as:

$$ x = \frac{1}{\sin 18^\circ} = \frac{4}{\sqrt{5} - 1} $$

### The Art of Rationalization

Now, we have an irrational number in the denominator. In the world of competitive exams, we never leave our answers in such a messy state.

We need to rationalize it. By multiplying both the numerator and the denominator by the conjugate, $(\sqrt{5} + 1)$, we transform the expression.

The denominator becomes $(\sqrt{5})^2 - (1)^2$, which is $5 - 1 = 4$. Suddenly, the denominator is no longer a source of anxiety; it is a simple integer.

The expression simplifies beautifully:

$$ x = \frac{4(\sqrt{5} + 1)}{4} $$

The fours cancel out, leaving us with the elegant result: $x = \sqrt{5} + 1$.

### The Algebraic Transformation

We are now at the final, most satisfying phase of our journey. We have the root $x = \sqrt{5} + 1$, and we need to turn this into a quadratic equation.

The secret, as I always tell my students, is to isolate the irrational part. If we shift the $1$ to the left side, we get $x - 1 = \sqrt{5}$.

Now, look at what happens when we square both sides:

$$ (x - 1)^2 = (\sqrt{5})^2 $$

The radical is gone! Expanding the left side using the identity $(a - b)^2 = a^2 - 2ab + b^2$, we get $x^2 - 2x + 1 = 5$.

Finally, bringing the $5$ to the left side gives us $x^2 - 2x + 1 - 5 = 0$, which simplifies to:

$$ x^2 - 2x - 4 = 0 $$

This is the quadratic equation we were looking for. It is a perfect match for our options.

See how the complexity melted away? Keep this technique in your toolkit, and you will find that even the most intimidating problems become manageable. **The final quadratic equation is $x^2 - 2x - 4 = 0$.**

Animated Solution for Mathematics - Trigonometry: $cosec 1 8^{\circ}$ is a root of the equation :

Select Answer:

Visualized Solution

The Objective

Recall $sin 1 8^{\circ}$

Relation: $cosec θ$ and $sin θ$

Rationalizing the Denominator

Simplifying the Fraction

Final Value of $x$

Isolating the Root

Squaring Both Sides

Expanding the LHS

Final Equation

The Sigma Insight: Trigonometric Ratios and Identities

The Elegant Dance of Trigonometry and Algebra

The Trigonometric Foundation

The Art of Rationalization

The Algebraic Transformation

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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; } cosec 18∘ is a root of the equation :

Select Answer:

Visualized Solution

The Objective

Recall .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin18∘

Relation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosec θ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ