Animated Solution for Mathematics - Trigonometry: The value of $\frac{\sqrt{3}\text{ cosec }20^{\circ }-\text{ sec }20^{\circ }}{\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{\circ }}$ is equal to

A Journey of Simplification

Imagine you are sitting in the examination hall. You turn the page, and there it is: a massive, intimidating fraction filled with $\text{cosec}$, $\sec$, and a long chain of $\cos$ terms.

Your heart might skip a beat. It looks like a chaotic mess of symbols. But here is the secret that every top-tier JEE aspirant knows: in mathematics, complexity is often just a mask for elegance.

This problem is not a monster; it is a puzzle waiting for you to find the right key.

The Numerator - The Art of Divide and Conquer

Let us look at the numerator: $\sqrt{3}\text{ cosec }20^{\circ }-\text{ sec }20^{\circ }$. The first instinct is often to panic, but we must remain calm. We have two different functions, $\text{cosec}$ and $\sec$, which we must translate into the universal language of trigonometry: $\sin$ and $\cos$.

We know that $\text{cosec }\theta =\frac{1}{\sin \theta }$ and $\text{sec }\theta =\frac{1}{\cos \theta }$. Substituting these, our expression becomes:

Now, we have a simple subtraction of fractions. By taking the least common multiple (LCM), we get:

Now, look closely at the numerator: $\sqrt{3}\cos 20^{\circ }-\sin 20^{\circ }$. This is the classic $a\cos \theta -b\sin \theta$ form. To simplify this, we multiply and divide by $\sqrt{(\sqrt{3}{)}^2+(1{)}^2}=2$.

This gives us:

Recognizing that $\frac{\sqrt{3}}{2}=\sin 60^{\circ }$ and $\frac{1}{2}=\cos 60^{\circ }$, the expression transforms into $2(\sin 60^{\circ }\cos 20^{\circ }-\cos 60^{\circ }\sin 20^{\circ })$. By the compound angle formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$, this collapses beautifully into:

The numerator is now a clean $2\sin 40^{\circ }$ divided by the denominator $\sin 20^{\circ }\cos 20^{\circ }$, which we know is $\frac{1}{2}\sin 40^{\circ }$. The $\sin 40^{\circ }$ terms cancel out, and we are left with $2/(1/2)=4$. The monster has been tamed!

The Denominator - The Beauty of Symmetry

Now, let us turn our attention to the denominator: $\cos 20^{\circ }\cos 40^{\circ }\cos 60^{\circ }\cos 80^{\circ }$. This is a beautiful, symmetric product. We know $\cos 60^{\circ }=\frac{1}{2}$, so we pull that out immediately.

We are left with $\frac{1}{2}(\cos 20^{\circ }\cos 40^{\circ }\cos 80^{\circ })$. Notice the angles: $20^{\circ },40^{\circ },80^{\circ }$. They are doubling!

This is the signature of the product formula:

Here, $\theta =20^{\circ }$ and $n=3$. Applying the formula, we get:

Since $\sin 160^{\circ }=\sin (180^{\circ }-20^{\circ })=\sin 20^{\circ }$, the expression simplifies to $\frac{\sin 20^{\circ }}{8\sin 20^{\circ }}=\frac{1}{8}$. Multiplying by the $\cos 60^{\circ }$ we pulled out earlier, the total denominator is:

The Grand Finale

We have conquered the numerator ($4$) and the denominator ($1/16$). The final step is simply:

Dividing by a fraction is the same as multiplying by its reciprocal, so $4\times 16=64$.

Look at that! From a chaotic expression, we have arrived at a simple, solid integer. This is the essence of JEE Advanced mathematics. It is not about brute force; it is about recognizing the patterns, applying the right tools, and trusting the process. You have the skills to solve this. Keep practicing, keep visualizing, and keep falling in love with the elegance of the math. The final answer is 64.