If `3cosectheta-7cot theta=3`, then find `3cottheta-7cosectheta=?`

Answer

To solve the problem, we start with the equation given:

1. **Given Equation**:
\[ 3 \csc \theta - 7 \cot \theta = 3 \]

2. **Rearranging the Equation**:
We can rearrange the equation to isolate one of the trigonometric functions:
\[ 3 \csc \theta = 3 + 7 \cot \theta \]

3. **Squaring Both Sides**:
Now, we square both sides of the equation:
\[ (3 \csc \theta)^2 = (3 + 7 \cot \theta)^2 \]
Expanding both sides:
\[ 9 \csc^2 \theta = 9 + 42 \cot \theta + 49 \cot^2 \theta \]

4. **Using Trigonometric Identities**:
We know that:
\[ \csc^2 \theta = 1 + \cot^2 \theta \]
Substituting this into the equation:
\[ 9(1 + \cot^2 \theta) = 9 + 42 \cot \theta + 49 \cot^2 \theta \]
Simplifying gives:
\[ 9 + 9 \cot^2 \theta = 9 + 42 \cot \theta + 49 \cot^2 \theta \]

5. **Rearranging Terms**:
Subtract \(9\) from both sides:
\[ 9 \cot^2 \theta = 42 \cot \theta + 49 \cot^2 \theta \]
Rearranging gives:
\[ 0 = 42 \cot \theta + 40 \cot^2 \theta \]

6. **Factoring**:
Factoring out \(\cot \theta\):
\[ \cot \theta (40 \cot \theta + 42) = 0 \]
This gives us two cases:
\[ \cot \theta = 0 \quad \text{or} \quad 40 \cot \theta + 42 = 0 \]

7. **Solving for \(\cot \theta\)**:
From \(40 \cot \theta + 42 = 0\):
\[ 40 \cot \theta = -42 \implies \cot \theta = -\frac{42}{40} = -\frac{21}{20} \]

8. **Finding \(3 \cot \theta - 7 \csc \theta\)**:
Now we need to find:
\[ 3 \cot \theta - 7 \csc \theta \]
We already have \(\cot \theta = -\frac{21}{20}\), so:
\[ 3 \cot \theta = 3 \left(-\frac{21}{20}\right) = -\frac{63}{20} \]

9. **Finding \(\csc \theta\)**:
Using the identity:
\[ \csc^2 \theta = 1 + \cot^2 \theta \]
We find \(\cot^2 \theta\):
\[ \cot^2 \theta = \left(-\frac{21}{20}\right)^2 = \frac{441}{400} \]
Thus:
\[ \csc^2 \theta = 1 + \frac{441}{400} = \frac{400 + 441}{400} = \frac{841}{400} \]
Therefore:
\[ \csc \theta = \sqrt{\frac{841}{400}} = \frac{29}{20} \]

10. **Calculating \(7 \csc \theta\)**:
\[ 7 \csc \theta = 7 \cdot \frac{29}{20} = \frac{203}{20} \]

11. **Final Calculation**:
Now substituting back:
\[ 3 \cot \theta - 7 \csc \theta = -\frac{63}{20} - \frac{203}{20} = -\frac{266}{20} = -\frac{133}{10} \]

Thus, the final answer is:
\[ \boxed{-\frac{133}{10}} \]