If `tan^(2)theta+cot^(2)theta=14`, then find `sectheta.cosectheta=?`

Answer

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

**Step 1: Write down the equation.**
Given:
\[ \tan^2 \theta + \cot^2 \theta = 14 \]

**Step 2: Substitute cotangent in terms of tangent.**
We know that:
\[ \cot \theta = \frac{1}{\tan \theta} \]
Thus, we can rewrite the equation as:
\[ \tan^2 \theta + \frac{1}{\tan^2 \theta} = 14 \]

**Step 3: Let \( x = \tan^2 \theta \).**
This gives us:
\[ x + \frac{1}{x} = 14 \]

**Step 4: Multiply through by \( x \) to eliminate the fraction.**
\[ x^2 + 1 = 14x \]

**Step 5: Rearrange the equation.**
\[ x^2 - 14x + 1 = 0 \]

**Step 6: Use the quadratic formula to solve for \( x \).**
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -14, c = 1 \):
\[ x = \frac{14 \pm \sqrt{(-14)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{14 \pm \sqrt{196 - 4}}{2} \]
\[ x = \frac{14 \pm \sqrt{192}}{2} \]
\[ x = \frac{14 \pm 8\sqrt{3}}{2} \]
\[ x = 7 \pm 4\sqrt{3} \]

**Step 7: Find \( \sec \theta \cdot \csc \theta \).**
Recall that:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Thus, we can find \( \sec^2 \theta \):
\[ \sec^2 \theta = 1 + x = 1 + (7 \pm 4\sqrt{3}) = 8 \pm 4\sqrt{3} \]

Now, we also need \( \csc^2 \theta \):
\[ \csc^2 \theta = 1 + \cot^2 \theta = 1 + \frac{1}{x} \]
Calculating \( \frac{1}{x} \):
\[ \frac{1}{x} = \frac{1}{7 \pm 4\sqrt{3}} \]
To simplify \( \csc^2 \theta \), we can rationalize:
\[ \csc^2 \theta = 1 + \frac{1}{7 \pm 4\sqrt{3}} = \frac{(7 \pm 4\sqrt{3}) + 1}{7 \pm 4\sqrt{3}} = \frac{8 \pm 4\sqrt{3}}{7 \pm 4\sqrt{3}} \]

**Step 8: Calculate \( \sec \theta \cdot \csc \theta \).**
Now, we can find:
\[ \sec \theta \cdot \csc \theta = \sqrt{\sec^2 \theta} \cdot \sqrt{\csc^2 \theta} = \sqrt{(8 \pm 4\sqrt{3}) \cdot \frac{8 \pm 4\sqrt{3}}{7 \pm 4\sqrt{3}}} \]
This simplifies to:
\[ \sec \theta \cdot \csc \theta = \pm 4 \]

Thus, the final answer is:
\[ \sec \theta \cdot \csc \theta = \pm 4 \]

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