The value of the expression √3 csc 20° - sec 20° is equal to (A) 2 (B) 2 sin 20° / sin 40° (C) 4 (D)

Q: Wait, why do we reject the negative root or why can't we just use a calculator to find the value?

In JEE Advanced, calculators are not permitted, and the goal is to develop your analytical intuition. Using a calculator would bypass the beautiful symmetry of the trigonometric identities we are exploring. We aren't just looking for a number; we are looking for the elegant path that leads to it.

Q: Why did we choose to multiply and divide by $2$ in the numerator? Is there a rule for this?

That is a fantastic question! We multiply by $2$ because we recognize the coefficients $\sqrt{3}$ and $1$ as part of the ratio $\frac{\sqrt{3}}{2}$ and $\frac{1}{2}$. These are the exact values of $\sin 60^\circ$ and $\cos 60^\circ$. Whenever you see an expression of the form $a \cos \theta + b \sin \theta$, multiplying by $\sqrt{a^2 + b^2}$ is a standard technique to force the expression into a sine or cosine addition formula.

Q: Is the vector method really necessary, or is the algebraic method better?

Both methods are powerful! The algebraic method is faster for quick calculations, but the vector method provides a deeper geometric insight. It shows that trigonometric identities are not just arbitrary rules; they are reflections of how vectors behave in space. Mastering both will make you a much more versatile problem solver.

Question

The value of the expression $\sqrt{3} \csc 20^\circ - \sec 20^\circ$ is equal to

Options:

(A) $2$

(B) $2 \sin 20^\circ / \sin 40^\circ$

(C) $4$

(D) $4 \sin 20^\circ / \sin 40^\circ$

Sigma Percentile · Accepted Answer

### The Beauty of the Hidden Angle: A Trigonometric Journey

Welcome, fellow explorers of the mathematical universe! Today, we are standing before a problem that might initially seem like a wall of confusion. We are asked to evaluate the expression $\sqrt{3} \csc 20^\circ - \sec 20^\circ$.

At first glance, the angle $20^\circ$ feels like an intruder. It is not one of our comfortable, standard angles like $30^\circ$, $45^\circ$, or $60^\circ$. But in the world of JEE Advanced, these 'awkward' angles are often just masks hiding a beautiful, symmetric truth.

### Phase 1: The Reciprocal Transformation

Our first step is to strip away the complexity. We know that $\csc 	heta$ is the reciprocal of $\sin 	heta$, and $\sec 	heta$ is the reciprocal of $\cos 	heta$. By applying these fundamental identities, our expression transforms into:

$$\frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}$$

Suddenly, the problem feels grounded. We are no longer dealing with obscure secants and cosecants; we are back in the familiar territory of sine and cosine.

To combine these, we find a common denominator, which is $\sin 20^\circ \cos 20^\circ$. The numerator becomes $\sqrt{3} \cos 20^\circ - \sin 20^\circ$. This is the heart of the problem.

### Phase 2: The Numerator's Secret

Look closely at that numerator: $\sqrt{3} \cos 20^\circ - \sin 20^\circ$. If you have been practicing your compound angle identities, you might feel a spark of recognition.

We have a coefficient of $\sqrt{3}$ and a coefficient of $1$. If we multiply and divide the entire numerator by $2$, we get:

$$2 \left( \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ ight)$$

Why did we do this? Because $\frac{\sqrt{3}}{2}$ is exactly $\sin 60^\circ$, and $\frac{1}{2}$ is exactly $\cos 60^\circ$.

Now, the expression inside the parentheses is $\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ$. This is the textbook definition of $\sin(A - B)$, where $A = 60^\circ$ and $B = 20^\circ$.

Thus, our numerator simplifies beautifully to $2 \sin(60^\circ - 20^\circ)$, which is $2 \sin 40^\circ$.

### Phase 3: The Denominator's Dance

Now, let us turn our attention to the denominator: $\sin 20^\circ \cos 20^\circ$. We recall the double-angle identity for sine: $\sin 2	heta = 2 \sin 	heta \cos 	heta$.

Our denominator is almost there, but it is missing a factor of $2$. So, we multiply and divide by $2$ again:

$$\frac{1}{2} (2 \sin 20^\circ \cos 20^\circ) = \frac{1}{2} \sin 40^\circ$$

### Phase 4: The Grand Finale

We have arrived at the final stage. Let us bring our simplified numerator and denominator back together:

$$\frac{2 \sin 40^\circ}{\frac{1}{2} \sin 40^\circ}$$

Notice the magic? The $\sin 40^\circ$ terms cancel out completely, leaving us with $2$ divided by $\frac{1}{2}$.

A quick calculation gives us $2 	imes 2 = 4$. The intimidating $20^\circ$ has vanished, leaving behind a clean, solid integer.

The final answer is **4**.

Animated Solution for Mathematics - Trigonometry: The value of the expression $3 csc 2 0^{\circ} - sec 2 0^{\circ}$ is equal to

Select Answer:

Visualized Solution

The Expression $3 csc 2 0^{\circ} - sec 2 0^{\circ}$

Reciprocal Identities

Common Denominator

Geometric Interpretation of the Numerator

Analyzing the Angle Between Vectors

Calculating the Dot Product (Numerator)

Algebraic Alternative for Numerator

Simplifying the Denominator

Final Division & Cancellation

Conclusion & Key Takeaways

The Sigma Insight: Trigonometric Functions of Compound Angles

The Beauty of the Hidden Angle

Phase 1

Phase 2

Phase 3

Phase 4

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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; } The value of the expression 3​csc20∘−sec20∘ is equal to