What is the value of `tan75^ @+ cot75^@` ?

Options

A. 2
B. 4
C. `2sqrt3`
D. `4sqrt3`

Answer

To find the value of \( \tan 75^\circ + \cot 75^\circ \), we can follow these steps:

### Step 1: Rewrite the angles
We know that:
\[ \tan 75^\circ = \tan(90^\circ - 15^\circ) = \cot 15^\circ \]
and
\[ \cot 75^\circ = \cot(90^\circ - 15^\circ) = \tan 15^\circ \]
Thus, we can rewrite the expression:
\[ \tan 75^\circ + \cot 75^\circ = \cot 15^\circ + \tan 15^\circ \]

### Step 2: Use the definitions of tangent and cotangent
Using the definitions:
\[ \cot 15^\circ = \frac{\cos 15^\circ}{\sin 15^\circ} \quad \text{and} \quad \tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} \]
We can substitute these into our expression:
\[ \cot 15^\circ + \tan 15^\circ = \frac{\cos 15^\circ}{\sin 15^\circ} + \frac{\sin 15^\circ}{\cos 15^\circ} \]

### Step 3: Find a common denominator
To combine these fractions, we need a common denominator:
\[ \cot 15^\circ + \tan 15^\circ = \frac{\cos^2 15^\circ + \sin^2 15^\circ}{\sin 15^\circ \cos 15^\circ} \]

### Step 4: Simplify the numerator
Using the Pythagorean identity:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
we can simplify the numerator:
\[ \cos^2 15^\circ + \sin^2 15^\circ = 1 \]
Thus, we have:
\[ \cot 15^\circ + \tan 15^\circ = \frac{1}{\sin 15^\circ \cos 15^\circ} \]

### Step 5: Use the double angle identity
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
So, we can express the denominator as:
\[ \sin 15^\circ \cos 15^\circ = \frac{1}{2} \sin 30^\circ \]
Since \( \sin 30^\circ = \frac{1}{2} \), we have:
\[ \sin 15^\circ \cos 15^\circ = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

### Step 6: Substitute back into the expression
Now substituting back, we get:
\[ \cot 15^\circ + \tan 15^\circ = \frac{1}{\frac{1}{4}} = 4 \]

### Final Answer
Thus, the value of \( \tan 75^\circ + \cot 75^\circ \) is:
\[ \boxed{4} \]