If `sintheta+costheta=pandsectheta+cosectheta=q`, then `q(p^(2)-1)=?`

Options

A. p
B. 2p
C. 3p
D. `2p^(2)`

Answer

To solve the problem, we need to find the value of \( q(p^2 - 1) \) given that \( \sin \theta + \cos \theta = p \) and \( \sec \theta + \csc \theta = q \).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ \sin \theta + \cos \theta = p \]
\[ \sec \theta + \csc \theta = q \]

2. **Express \( q \) in terms of \( \sin \theta \) and \( \cos \theta \):**
\[ q = \sec \theta + \csc \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} \]
To combine these fractions, find a common denominator:
\[ q = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} \]

3. **Substitute \( p \) into the equation for \( q \):**
Since \( \sin \theta + \cos \theta = p \), we can write:
\[ q = \frac{p}{\sin \theta \cos \theta} \]

4. **Now, express \( p^2 - 1 \):**
\[ p^2 - 1 = (\sin \theta + \cos \theta)^2 - 1 \]
Expanding this:
\[ p^2 - 1 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1 \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ p^2 - 1 = 1 + 2\sin \theta \cos \theta - 1 = 2\sin \theta \cos \theta \]

5. **Now we can find \( q(p^2 - 1) \):**
Substitute \( p^2 - 1 \) into the expression:
\[ q(p^2 - 1) = q \cdot 2\sin \theta \cos \theta \]
Substitute \( q = \frac{p}{\sin \theta \cos \theta} \):
\[ q(p^2 - 1) = \left(\frac{p}{\sin \theta \cos \theta}\right) \cdot 2\sin \theta \cos \theta \]
The \( \sin \theta \cos \theta \) cancels out:
\[ q(p^2 - 1) = 2p \]

6. **Final answer:**
\[ q(p^2 - 1) = 2p \]