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*}$