For the following two (02) items: Let $\sin\theta+\cos\theta = p$ and $\sec\theta+\text{cosec}\theta = q$, where $p \neq 1$.
Solving the Trigonometric Relation Between p and q
This problem asks us to find a relationship between two given expressions involving trigonometric functions. We are given:
- \(p = \sin\theta + \cos\theta\)
- \(q = \sec\theta + \text{cosec}\theta\)
- A condition that \(p \neq 1\).
Our goal is to manipulate these equations algebraically to arrive at one of the given options, specifically the relation between \(p\) and \(q\).
Step-by-Step Derivation
Step 1: Express \(p^2\) in terms of \(\sin\theta\) and \(\cos\theta\)
Let's start by squaring the expression for \(p\):
\(p^2 = (\sin\theta + \cos\theta)^2\)
Expanding the square using the formula \((a+b)^2 = a^2 + b^2 + 2ab\):
\(p^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta\)
We know the fundamental trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\). Substituting this:
\(p^2 = 1 + 2\sin\theta\cos\theta\)
Now, we can isolate the term \(2\sin\theta\cos\theta\):
\(2\sin\theta\cos\theta = p^2 - 1\)
This gives us an expression for the product \(\sin\theta\cos\theta\) in terms of \(p\):
\(\sin\theta\cos\theta = \frac{p^2 - 1}{2}\)
The condition \(p \neq 1\) ensures that \(p^2 - 1 \neq 0\), which means \(\sin\theta\cos\theta \neq 0\). This is important because \(\sin\theta\) and \(\cos\theta\) appear in the denominators of \(\sec\theta\) and \(\text{cosec}\theta\), respectively.
Step 2: Express \(q\) in terms of \(\sin\theta\) and \(\cos\theta\)
Now let's work with the expression for \(q\). We can rewrite \(\sec\theta\) as \(\frac{1}{\cos\theta}\) and \(\text{cosec}\theta\) as \(\frac{1}{\sin\theta}\):
\(q = \frac{1}{\cos\theta} + \frac{1}{\sin\theta}\)
To combine these fractions, we find a common denominator, which is \(\sin\theta\cos\theta\):
\(q = \frac{\sin\theta}{\sin\theta\cos\theta} + \frac{\cos\theta}{\sin\theta\cos\theta}\)
\(q = \frac{\sin\theta + \cos\theta}{\sin\theta\cos\theta}\)
Step 3: Substitute expressions involving \(p\) into the expression for \(q\)
We have already found expressions for the numerator \((\sin\theta + \cos\theta)\) and the denominator \((\sin\theta\cos\theta)\) in terms of \(p\):
- Numerator: \(\sin\theta + \cos\theta = p\)
- Denominator: \(\sin\theta\cos\theta = \frac{p^2 - 1}{2}\)
Substitute these into the equation for \(q\):
\(q = \frac{p}{\left(\frac{p^2 - 1}{2}\right)}\)
To simplify this complex fraction, we multiply the numerator by the reciprocal of the denominator:
\(q = p \times \frac{2}{p^2 - 1}\)
\(q = \frac{2p}{p^2 - 1}\)
Step 4: Rearrange the equation to match the options
The current equation is \(q = \frac{2p}{p^2 - 1}\). We need to rearrange this to match one of the provided options. Let's multiply both sides by \((p^2 - 1)\) to clear the denominator:
\(q(p^2 - 1) = 2p\)
This equation directly matches option 2. We can also write it as \(2p = q(p^2 - 1)\).
Final Relation Between p and q
The derived relation between \(p\) and \(q\) is \(2p = q(p^2 - 1)\). This identity holds true given the initial definitions of \(p\) and \(q\) and the condition \(p \neq 1\).
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 $\tan\theta + \cot\theta$ 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 ?
Consider the following statements :
I. \(\tan 50^\circ - \cot 50^\circ\) is positive
II. \(\cot 25^\circ - \tan 25^\circ\) is negative
Which of the statements is/are correct?
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