$p = \frac{\sin \theta}{1 + \cos \theta + \sin \theta} \text{ and } q = \frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}$
Expression Setup
We need to find the value of the expression \(\left(p+\frac{1}{q}\right)\left(q+\frac{1}{p}\right)\). The given expressions for \(p\) and \(q\) are:
\(p = \frac{\sin \theta}{1 + \cos \theta + \sin \theta}\)
\(q = \frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}\)
Simplify p
We use the substitution \(t = \tan\left(\frac{\theta}{2}\right)\). This implies: \(\sin \theta = \frac{2t}{1+t^2}\) and \(\cos \theta = \frac{1-t^2}{1+t^2}\).
Substituting these into the expression for \(p\): \(p = \frac{\frac{2t}{1+t^2}}{1 + \frac{1-t^2}{1+t^2} + \frac{2t}{1+t^2}}\)
To simplify, multiply the numerator and denominator by \(1+t^2\): \(p = \frac{2t}{(1+t^2) + (1-t^2) + 2t} = \frac{2t}{2 + 2t} = \frac{t}{1+t}\).
Simplify q
Using the same substitutions for \(\sin \theta\) and \(\cos \theta\) in the expression for \(q\):
\(q = \frac{1 + \frac{2t}{1+t^2}}{1 + \frac{2t}{1+t^2} - \frac{1-t^2}{1+t^2}}\)
Multiply the numerator and denominator by \(1+t^2\): \(q = \frac{(1+t^2) + 2t}{(1+t^2) + 2t - (1-t^2)} = \frac{(1+t)^2}{1+t^2+2t-1+t^2} = \frac{(1+t)^2}{2t+2t^2} = \frac{(1+t)^2}{2t(1+t)}\).
Simplifying further, we get: \(q = \frac{1+t}{2t}\).
p and q Reciprocals
We calculate the reciprocal terms required for the final expression:
- \(\frac{1}{p} = \frac{1}{\frac{t}{1+t}} = \frac{1+t}{t}\)
- \(\frac{1}{q} = \frac{1}{\frac{1+t}{2t}} = \frac{2t}{1+t}\)
Evaluate (p + 1/q)
Now, we find the value of the first bracket term \(p+\frac{1}{q}\):
\(p+\frac{1}{q} = \frac{t}{1+t} + \frac{2t}{1+t} = \frac{t+2t}{1+t} = \frac{3t}{1+t}\).
Evaluate (q + 1/p)
Next, we find the value of the second bracket term \(q+\frac{1}{p}\):
\(q+\frac{1}{p} = \frac{1+t}{2t} + \frac{1+t}{t} = \frac{1+t}{2t} + \frac{2(1+t)}{2t} = \frac{(1+t) + 2(1+t)}{2t} = \frac{3(1+t)}{2t}\).
Multiply Bracket Terms
Finally, we multiply the results of the two bracket terms:
\(\left(p+\frac{1}{q}\right)\left(q+\frac{1}{p}\right) = \left(\frac{3t}{1+t}\right) \left(\frac{3(1+t)}{2t}\right)\)
\(= \frac{3t \cdot 3(1+t)}{(1+t) \cdot 2t} = \frac{9t(1+t)}{2t(1+t)}\).
Final Value
By cancelling the common terms \(t\) and \((1+t)\) (assuming \(t \neq 0\) and \(t \neq -1\)), the expression simplifies to:
\(\frac{9}{2}\).
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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