Question

The value of cos 1200° + tan 1485° is

• A. \(\frac{1}{2}\)
• B. \(\frac{3}{2}\)
• C. \(-\frac{3}{2}\)
• D. \(-\frac{1}{2}\)

Answer 1

The Correct Option is A

Let's break down the trigonometric expressions:

1. For \( \cos 120^\circ \): \( \cos 120^\circ = -\frac{1}{2} \), since \( \cos 120^\circ \) lies in the second quadrant where cosine is negative.
2. For \( \tan 1485^\circ \): First, reduce the angle \( 1485^\circ \) to an angle between 0° and 360°: \[ 1485^\circ \div 360^\circ = 4 \text{ full circles with remainder } 1485 - 4 \times 360 = 1485 - 1440 = 45^\circ. \] Thus, \( \tan 1485^\circ = \tan 45^\circ = 1 \), since \( \tan 45^\circ = 1 \).

So, we have: \[ \cos 120^\circ + \tan 1485^\circ = -\frac{1}{2} + 1 = \frac{1}{2}. \]

The correct answer is (A) : \(\frac 12.\)

Answer 2

We need to evaluate the expression \( \cos(1200^\circ) + \tan(1485^\circ) \).

Step 1: Evaluate \( \cos(1200^\circ) \) 

The cosine function has a period of 360°. We can find an equivalent angle within the range [0°, 360°) by subtracting multiples of 360°.

\[ 1200^\circ = n \times 360^\circ + \theta \]

Divide 1200 by 360:

\[ \frac{1200}{360} \approx 3.33 \]

Calculate \( 3 \times 360^\circ = 1080^\circ \).

\[ 1200^\circ = 1080^\circ + 120^\circ = 3 \times 360^\circ + 120^\circ \]

Therefore,

\[ \cos(1200^\circ) = \cos(3 \times 360^\circ + 120^\circ) = \cos(120^\circ) \]

Now, evaluate \( \cos(120^\circ) \). The angle 120° is in the second quadrant.

In the second quadrant, cosine is negative.

The reference angle is \( 180^\circ - 120^\circ = 60^\circ \).

\[ \cos(120^\circ) = -\cos(60^\circ) \]

We know that \( \cos(60^\circ) = \frac{1}{2} \).

\[\cos(1200^\circ) = \cos(120^\circ) = -\frac{1}{2}\]

Step 2: Evaluate \( \tan(1485^\circ) \)

The tangent function has a period of 180° (or 360°). We can use 360°.

\[ 1485^\circ = n \times 360^\circ + \theta \]

Divide 1485 by 360:

\[ \frac{1485}{360} \approx 4.125 \]

Calculate \( 4 \times 360^\circ = 1440^\circ \).

\[ 1485^\circ = 1440^\circ + 45^\circ = 4 \times 360^\circ + 45^\circ \]

Therefore,

\[ \tan(1485^\circ) = \tan(4 \times 360^\circ + 45^\circ) = \tan(45^\circ) \]

We know that \( \tan(45^\circ) = 1 \).

\[\tan(1485^\circ) = 1\]

Step 3: Calculate the sum

\[ \cos(1200^\circ) + \tan(1485^\circ) = -\frac{1}{2} + 1 \]

\[ = -\frac{1}{2} + \frac{2}{2} \]

\[ = \frac{-1 + 2}{2} \]

\[= \frac{1}{2}\]

The value of the expression is \( \frac{1}{2} \).

Comparing this result with the given options:

• \( \frac{1}{2} \)
• \( \frac{3}{2} \)
• \( -\frac{3}{2} \)
• \( -\frac{1}{2} \)

The correct option is \( \frac{1}{2} \).