$ p + q \cot \theta = 3 \cosec \theta $ and $ q - p \cot \theta = 2 \cosec \theta $
Trigonometric Equation Analysis
We are given the equations:
- \(p + q \cot \theta = 3 \cosec \theta\) (1)
- \(q - p \cot \theta = 2 \cosec \theta\) (2)
The objective is to find \(\tan \theta\).
Eliminating \(\cosec \theta\)
Multiply Equation (1) by 2 and Equation (2) by 3 to standardize the \(\cosec \theta\) term on the right side:
- \(2 \times (p + q \cot \theta) = 2 \times 3 \cosec \theta \implies 2p + 2q \cot \theta = 6 \cosec \theta\)
- \(3 \times (q - p \cot \theta) = 3 \times 2 \cosec \theta \implies 3q - 3p \cot \theta = 6 \cosec \theta\)
Equating the left-hand sides since the right-hand sides are equal:
\(2p + 2q \cot \theta = 3q - 3p \cot \theta\)
Solving for \(\cot \theta\)
Rearrange the equation to group \(\cot \theta\) terms on one side and constants on the other:
\(2q \cot \theta + 3p \cot \theta = 3q - 2p\)
Factor out \(\cot \theta\):
\(\cot \theta (2q + 3p) = 3q - 2p\)
Isolate \(\cot \theta\):
\(\cot \theta = \frac{3q - 2p}{2q + 3p}\)
Calculating \(\tan \theta\)
Using the reciprocal identity \(\tan \theta = \frac{1}{\cot \theta}\):
\(\tan \theta = \frac{1}{\frac{3q - 2p}{2q + 3p}}\)
\(\tan \theta = \frac{2q + 3p}{3q - 2p}\)
The final answer is \(\boxed{\frac{(2q+3p)}{(3q+2p)}}\)
Similar Questions
If \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} = \text{cosec}\theta\), then what is \(\sin\theta\) equal to?
- If \(2 \tan\theta = \sec^2\theta-2\), where \(0 < \theta < \pi /2\), then what is \(\cot\theta\) equal to?
- What is \(\frac{\tan^2 \alpha}{\tan^2 \beta}\) equal to?
- If \(\alpha\) and \(\beta\) are complementary angles, then which one of the following is correct?
- What is \(\sec\theta\) equal to?
- What is \(\tan\theta\) equal to?
- What is the relation between \(p\) and \(q\)?
- What is $\tan\theta + \cot\theta$ equal to?
- Consider the following statements :
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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