Animated Solution for Mathematics - Trigonometry: If $\tan 15^{\circ }+\frac{1}{\tan 75^{\circ }}+\frac{1}{\tan 105^{\circ }}+\tan 195^{\circ }=2a$, then the value of $a+\frac{1}{a}$ is:

Welcome, fellow traveler of the JEE journey. Today, we are standing before a trigonometric expression that might look like a tangled thicket of angles:

At first glance, it is easy to feel overwhelmed. But in the world of JEE Advanced, complexity is often just a mask for hidden symmetry. Our goal is to find $a+\frac{1}{a}$. Let us peel back the layers.

The Angle Hunt

Look at the angles: $15^{\circ },75^{\circ },105^{\circ },195^{\circ }$. They are all cousins of $15^{\circ }$.

Our strategy is simple: convert everything into a function of $15^{\circ }$. First, recall the standard value $\tan 15^{\circ }=2-\sqrt{3}$. This is a fundamental building block you should keep in your mental toolkit.

The Reduction Dance

Take the second term, $\frac{1}{\tan 75^{\circ }}$. We know that $\frac{1}{\tan \theta }=\cot \theta$, so this becomes $\cot 75^{\circ }$.

Since $75^{\circ }=90^{\circ }-15^{\circ }$, we use the identity $\cot (90^{\circ }-\theta )=\tan \theta$. Thus, $\cot 75^{\circ }=\tan 15^{\circ }$. The term simplifies instantly.

Next, the third term: $\frac{1}{\tan 105^{\circ }}=\cot 105^{\circ }$. We can write $105^{\circ }$ as $180^{\circ }-75^{\circ }$.

In the second quadrant, $\cot$ is negative, so $\cot (180^{\circ }-75^{\circ })=-\cot 75^{\circ }$. And since we already know $\cot 75^{\circ }=\tan 15^{\circ }$, this term becomes $-\tan 15^{\circ }$.

Finally, the fourth term: $\tan 195^{\circ }$. We write this as $\tan (180^{\circ }+15^{\circ })$.

In the third quadrant, tangent is positive, so $\tan (180^{\circ }+15^{\circ })=\tan 15^{\circ }$. Everything is now in terms of $\tan 15^{\circ }$.

The Algebraic Resolution

Substitute these back into the original equation:

The terms $\tan 15^{\circ }$ and $-\tan 15^{\circ }$ cancel out, leaving us with $2\tan 15^{\circ }=2a$. Dividing by $2$, we find $a=\tan 15^{\circ }=2-\sqrt{3}$.

Now, for the final act: $a+\frac{1}{a}$. We have $a=2-\sqrt{3}$.

Then $\frac{1}{a}=\frac{1}{2-\sqrt{3}}$. Rationalizing the denominator by multiplying the numerator and denominator by $2+\sqrt{3}$, we get:

Adding them together, $(2-\sqrt{3})+(2+\sqrt{3})=4$. The symmetry is complete. We started with a complex expression and ended with a clean, integer result of 4.