Question

If $\tan 15^{\circ}+\frac{1}{\tan 75^{\circ}}+\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2 a$, then the value of $\left(a+\frac{1}{a}\right)$ is :

• A. 4
• B. 2
• C. $4-2 \sqrt{3}$
• D. $5-\frac{3}{2} \sqrt{3}$

Answer 1

The Correct Option is A

To solve this problem, we start by examining the given expression:

\(\tan 15^{\circ}+\frac{1}{\tan 75^{\circ}}+\frac{1}{\tan 105^{\circ}}+\tan 195^{\circ}=2a\)

First, we use the identity \(\tan(180^\circ + \theta) = \tan \theta\). Therefore, we have:

• \(\tan 195^{\circ} = \tan (180^\circ + 15^\circ) = \tan 15^{\circ}\)

The expression simplifies to:

\(\tan 15^{\circ} + \frac{1}{\tan 75^{\circ}} + \frac{1}{\tan 105^{\circ}} + \tan 15^{\circ} = 2a\)

Combining similar terms, we get:

\(2\tan 15^{\circ} + \frac{1}{\tan 75^{\circ}} + \frac{1}{\tan 105^{\circ}} = 2a\)

Using the identity \(\tan(90^\circ - \theta) = \cot \theta\), we have:

• \(\tan 75^{\circ} = \cot 15^\circ = \frac{1}{\tan 15^\circ}\)
• \(\tan 105^{\circ} = \cot 75^\circ = \frac{1}{\tan 75^\circ} = \tan 15^\circ\)

Thus, \(\frac{1}{\tan 75^{\circ}} = \tan 15^{\circ}\) and \(\frac{1}{\tan 105^{\circ}} = \frac{1}{\tan (90^\circ + 15^\circ)} = -\tan 15^{\circ}\).

Substituting these in, the expression becomes:

\(2\tan 15^{\circ} + \tan 15^{\circ} - \tan 15^{\circ} = 2a\)

Simplifying gives:

\(2\tan 15^{\circ} = 2a\)

Dividing both sides by 2, we find:

\(a = \tan 15^{\circ}\)

We now focus on finding the value of \(a + \frac{1}{a}\):

\(a + \frac{1}{a} = \tan 15^{\circ} + \frac{1}{\tan 15^{\circ}}\)

The identity for \(\tan\) and \(\cot\) implies:

\(a + \frac{1}{a} = \tan 15^{\circ} + \cot 15^{\circ} = \tan 15^{\circ} + \frac{1}{\tan 15^{\circ}} = 2\cot 30^{\circ} = 2\sqrt{3}\)

Thus, the calculated expression simplifies to \(4\).

Therefore, the value of \((a + \frac{1}{a})\) is:

Option: 4