Let `l= sin theta, m=cos theta and n=tan theta`.
Q. If `theta=7` radian, then :

Options

A. `l+m gt 0`
B. `l+m lt 0`
C. `l+m=0`
D. nothing can be said

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Convert Radians to Degrees
We need to convert the angle \( \theta = 7 \) radians into degrees. The formula for converting radians to degrees is:

\[ \text{Degrees} = \text{Radians} \times \frac{180}{\pi} \]

Substituting \( \theta = 7 \):

\[ \text{Degrees} = 7 \times \frac{180}{\pi} \]

Using \( \pi \approx 3.14 \):

\[ \text{Degrees} \approx 7 \times \frac{180}{3.14} \approx 7 \times 57.2958 \approx 401.07 \]

### Step 2: Simplify the Angle
Since trigonometric functions are periodic with a period of \( 360^\circ \), we can simplify \( 401.07^\circ \) by subtracting \( 360^\circ \):

\[ 401.07 - 360 = 41.07^\circ \]

Thus, we can approximate \( \theta \) as \( 41^\circ \).

### Step 3: Calculate \( l \), \( m \), and \( n \)
Now we can find \( l \), \( m \), and \( n \):

- \( l = \sin(41^\circ) \)
- \( m = \cos(41^\circ) \)
- \( n = \tan(41^\circ) \)

### Step 4: Analyze the Values of \( l \) and \( m \)
Since both \( 41^\circ \) is an acute angle, we know:

- \( \sin(41^\circ) > 0 \)
- \( \cos(41^\circ) > 0 \)

Therefore, both \( l \) and \( m \) are positive.

### Step 5: Conclusion
Since both \( l \) and \( m \) are positive, their sum \( l + m \) will also be positive:

\[ l + m > 0 \]

Thus, the statement that \( l + m \) is greater than \( 0 \) is true.

### Final Answer
The correct option is that \( l + m > 0 \).

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