$p = \frac{\sin \theta}{1 + \cos \theta + \sin \theta} \text{ and } q = \frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}$
\(2pq - 1 = 0\)
To solve the given problem, we need to determine which mathematical expression among the options is correct based on the provided values of \( p \) and \( q \):
Given: \(p = \frac{\sin \theta}{1 + \cos \theta + \sin \theta}\) and \(q = \frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}\)
- Substituting \( p \) and \( q \) into the expressions:
- \(p - q = \frac{\sin \theta}{1 + \cos \theta + \sin \theta} - \frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}\)
- \(pq - 1 = \left(\frac{\sin \theta}{1 + \cos \theta + \sin \theta}\right) \left(\frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}\right) - 1\)
- \(pq - 2 = \left(\frac{\sin \theta}{1 + \cos \theta + \sin \theta}\right) \left(\frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}\right) - 2\)
- \(2pq - 1 = 2\left(\frac{\sin \theta}{1 + \cos \theta + \sin \theta}\right) \left(\frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}\right) - 1\)
- Calculate \( pq \):
First, find the product \( pq \):
\(pq = \left(\frac{\sin \theta}{1 + \cos \theta + \sin \theta}\right) \left(\frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}\right)\)
By simplifying the expression \( pq = \frac{\sin \theta (1 + \sin \theta)}{(1 + \cos \theta + \sin \theta)(1 + \sin \theta - \cos \theta)} \).
Notice that the denominator may be simplified using algebraic identities.
After simplification, we find that \(2pq - 1 = 0\) turns out to be true.
- Verification:
Substitute and simplify the expression to check the validity:
This involves simplification of trigonometric identities which leads to verifying:
\(2pq = 1 \implies \frac{2 \sin \theta (1 + \sin \theta)}{(1 + \cos \theta + \sin \theta)(1 + \sin \theta - \cos \theta)} = 1\)
Upon finding these values and verifying all parts, we conclude:
- Conclusion:
Based on the calculations and verification, the correct answer is indeed:
Option: \(2pq - 1 = 0\)
Similar Questions
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1. In a triangle ABC, if \(\sin A + \sin B + \sin C = \frac{3\sqrt{3}}{2}\), then the triangle can be equilateral.
2. In a triangle ABC, if \(\cos A + \cos B + \cos C = \frac{3}{2}\), then the triangle can be equilateral.
Which of the statements given above is/are correct? What is \((1 + \tan\alpha \tan\beta)^2 + (\tan\alpha - \tan\beta)^2\) equal to ?
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