Which of the following pairs may represent the eccentricities of two conjugate hyperbolas, for `alpha in (0,pi//2)`?
a.`sin theta, cos theta` b.`tan theta, cot theta` c.`sec theta, "cosec"theta` d.`1+sintheta,1+cos theta`

Options

A. `sin theta, cos theta`
B. `tan theta, cot theta`
C. `sec theta, "cosec"theta`
D. `1+sintheta,1+cos theta`

Explanation

To determine which of the given pairs may represent the eccentricities of two conjugate hyperbolas, we will use the relationship between the eccentricities \( E_1 \) and \( E_2 \) of the hyperbolas. The relationship states that:

\[ \frac{1}{E_1^2} + \frac{1}{E_2^2} = 1 \]

We will analyze each option step by step.

### Step 1: Analyze Option A - \( \sin \theta, \cos \theta \)

Let:
- \( E_1 = \sin \theta \)
- \( E_2 = \cos \theta \)

Now, substituting into the relationship:

\[ \frac{1}{(\sin \theta)^2} + \frac{1}{(\cos \theta)^2} = \frac{\cos^2 \theta + \sin^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} \]

This does not equal 1, so this option is **not valid**.

### Step 2: Analyze Option B - \( \tan \theta, \cot \theta \)

Let:
- \( E_1 = \tan \theta \)
- \( E_2 = \cot \theta \)

Substituting into the relationship:

\[ \frac{1}{(\tan \theta)^2} + \frac{1}{(\cot \theta)^2} = \frac{\cot^2 \theta + \tan^2 \theta}{\tan^2 \theta \cot^2 \theta} = \frac{\frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta}}{1} = \frac{1}{\sin^2 \theta \cos^2 \theta} \]

This does not equal 1, so this option is **not valid**.

### Step 3: Analyze Option C - \( \sec \theta, \csc \theta \)

Let:
- \( E_1 = \sec \theta \)
- \( E_2 = \csc \theta \)

Substituting into the relationship:

\[ \frac{1}{(\sec \theta)^2} + \frac{1}{(\csc \theta)^2} = \cos^2 \theta + \sin^2 \theta = 1 \]

This holds true, so this option is **valid**.

### Step 4: Analyze Option D - \( 1 + \sin \theta, 1 + \cos \theta \)

Let:
- \( E_1 = 1 + \sin \theta \)
- \( E_2 = 1 + \cos \theta \)

Substituting into the relationship:

\[ \frac{1}{(1 + \sin \theta)^2} + \frac{1}{(1 + \cos \theta)^2} \]

This expression does not simplify to 1, so this option is **not valid**.

### Conclusion

The only valid pair of eccentricities for two conjugate hyperbolas is from Option C: \( \sec \theta \) and \( \csc \theta \).

### Final Answer:
**Option C: \( \sec \theta, \csc \theta \)**

---