For the following two (02) items: Let $\sin\theta+\cos\theta = p$ and $\sec\theta+\text{cosec}\theta = q$, where $p \neq 1$.
Solving Trigonometric Expression tanθ + cotθ
This solution explains how to find the value of the trigonometric expression \(\tan\theta + \cot\theta\)$ given specific relationships involving \(\sin\theta\), \(\cos\theta\), \(\sec\theta\),\)$ and \(\text{cosec}\theta\)$.
Understanding the Given Information
We are provided with two equations:
\(\sin\theta + \cos\theta = p\)$\(\sec\theta + \text{cosec}\theta = q\)$
We are also given the condition that \(\_p \neq 1\)$.
Simplifying the Target Expression
The expression we need to evaluate is \(\tan\theta + \cot\theta\)$. Let's rewrite this in terms of \(\sin\theta\)$ and \(\cos\theta\)$:
\begin{equation*} \tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \end{equation*}$
To add these fractions, we find a common denominator:
\begin{equation*} \frac{\sin^2\theta + \cos^2\theta}{\sin\theta \cos\theta} \end{equation*}$
Using the fundamental trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\)$, the expression simplifies to:
\begin{equation*} \frac{1}{\sin\theta \cos\theta} \end{equation*}$
So, our goal now is to find the value of \(\sin\theta \cos\theta\)$ using the given information.
Finding the Value of \( \sin\theta \cos\theta \)
Let's use the first given equation, \(\sin\theta + \cos\theta = p\)$. We can square both sides of this equation:
\begin{equation*} (\sin\theta + \cos\theta)^2 = p^2 \end{equation*}$
Expanding the left side gives:
\begin{equation*} \sin^2\theta + \cos^2\theta + 2\sin\theta \cos\theta = p^2 \end{equation*}$
Substituting \(\sin^2\theta + \cos^2\theta = 1\)$:
\begin{equation*} 1 + 2\sin\theta \cos\theta = p^2 \end{equation*}$
Rearranging to solve for \(\sin\theta \cos\theta\)$:
\begin{equation*} 2\sin\theta \cos\theta = p^2 - 1 \end{equation*}$
\begin{equation*} \sin\theta \cos\theta = \frac{p^2 - 1}{2} \end{equation*}$
Now let's use the second given equation, \(\sec\theta + \text{cosec}\theta = q\)$. Rewrite \(\sec\theta\)$ and \(\text{cosec}\theta\)$ in terms of \(\sin\theta\)$ and \(\cos\theta\)$:
\begin{equation*} \frac{1}{\cos\theta} + \frac{1}{\sin\theta} = q \end{equation*}$
Combine the fractions on the left side:
\begin{equation*} \frac{\sin\theta + \cos\theta}{\sin\theta \cos\theta} = q \end{equation*}$
We know that \(\sin\theta + \cos\theta = p\)$. Substitute this into the equation:
\begin{equation*} \frac{p}{\sin\theta \cos\theta} = q \end{equation*}$
Now, solve for \(\sin\theta \cos\theta\)$:
\begin{equation*} \sin\theta \cos\theta = \frac{p}{q} \end{equation*}$
Calculating the Final Value
We established earlier that \(\tan\theta + \cot\theta = \frac{1}{\sin\theta \cos\theta}\)$.
Using the value \(\sin\theta \cos\theta = \frac{p}{q}\)$ derived from the second equation:
\begin{equation*} \tan\theta + \cot\theta = \frac{1}{\frac{p}{q}} \end{equation*}$
Simplifying this compound fraction gives:
\begin{equation*} \tan\theta + \cot\theta = \frac{q}{p} \end{equation*}$
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 \(q\)?
- 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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