Solving the Trigonometry Expression: \( \frac{p^2 - 1}{p^2 + 1} \)

We are given the equation \( \text{cosec } \theta + \text{cot } \theta = p \). We need to find the value of the expression \( \frac{p^2 - 1}{p^2 + 1} \).

Let's use a fundamental trigonometric identity involving cosecant and cotangent.

Key Trigonometric Identity

Recall the identity:

\( \text{cosec}^2 \theta - \text{cot}^2 \theta = 1 \)

This identity is a difference of squares, which can be factored as:

\( (\text{cosec } \theta - \text{cot } \theta)(\text{cosec } \theta + \text{cot } \theta) = 1 \)

Using the Given Information

We are given that \( \text{cosec } \theta + \text{cot } \theta = p \). Substituting this into the factored identity:

\( (\text{cosec } \theta - \text{cot } \theta)(p) = 1 \)

From this, we can find an expression for \( \text{cosec } \theta - \text{cot } \theta \):

\( \text{cosec } \theta - \text{cot } \theta = \frac{1}{p} \)

Setting up a System of Equations

Now we have two useful equations:

1. \( \text{cosec } \theta + \text{cot } \theta = p \)
2. \( \text{cosec } \theta - \text{cot } \theta = \frac{1}{p} \)

We can solve this system for \( \text{cosec } \theta \) and \( \text{cot } \theta \) in terms of \( p \).

Adding the Equations

Add equation (1) and equation (2):

\( (\text{cosec } \theta + \text{cot } \theta) + (\text{cosec } \theta - \text{cot } \theta) = p + \frac{1}{p} \)

\( 2\text{cosec } \theta = \frac{p^2 + 1}{p} \)

So,

\( \text{cosec } \theta = \frac{p^2 + 1}{2p} \)

Subtracting the Equations

Subtract equation (2) from equation (1):

\( (\text{cosec } \theta + \text{cot } \theta) - (\text{cosec } \theta - \text{cot } \theta) = p - \frac{1}{p} \)

\( 2\text{cot } \theta = \frac{p^2 - 1}{p} \)

So,

\( \text{cot } \theta = \frac{p^2 - 1}{2p} \)

Finding the Value of \( \frac{p^2 - 1}{p^2 + 1} \)

We have found expressions for \( \text{cosec } \theta \) and \( \text{cot } \theta \) in terms of \( p \):

• \( \text{cosec } \theta = \frac{p^2 + 1}{2p} \)
• \( \text{cot } \theta = \frac{p^2 - 1}{2p} \)

We know that \( \text{cot } \theta = \frac{\text{cos } \theta}{\text{sin } \theta} \) and \( \text{cosec } \theta = \frac{1}{\text{sin } \theta} \). Therefore, \( \text{cos } \theta = \text{cot } \theta \times \text{sin } \theta \).

Substituting the expressions in terms of \( p \):

\( \text{cos } \theta = \left(\frac{p^2 - 1}{2p}\right) \times \left(\frac{1}{\text{cosec } \theta}\right) \)

\( \text{cos } \theta = \left(\frac{p^2 - 1}{2p}\right) \times \left(\frac{1}{\frac{p^2 + 1}{2p}}\right) \)

\( \text{cos } \theta = \left(\frac{p^2 - 1}{2p}\right) \times \left(\frac{2p}{p^2 + 1}\right) \)

We can cancel the \( 2p \) terms:

\( \text{cos } \theta = \frac{p^2 - 1}{p^2 + 1} \)

Thus, the value of \( \frac{p^2 - 1}{p^2 + 1} \) is \( \text{cos } \theta \).

Summary of Results

Our calculation shows that \( \frac{p^2 - 1}{p^2 + 1} \) is equal to \( \text{cos } \theta \).

Revision Table: Key Trigonometry Concepts

Additional Information: Alternative Approach

Another way to approach this problem is to express \( p \) directly in terms of \( \text{sin } \theta \) and \( \text{cos } \theta \) and then substitute it into the expression \( \frac{p^2 - 1}{p^2 + 1} \).

Given \( p = \text{cosec } \theta + \text{cot } \theta \), we have:

\( p = \frac{1}{\text{sin } \theta} + \frac{\text{cos } \theta}{\text{sin } \theta} = \frac{1 + \text{cos } \theta}{\text{sin } \theta} \)

Now, calculate \( p^2 \):

\( p^2 = \left(\frac{1 + \text{cos } \theta}{\text{sin } \theta}\right)^2 = \frac{(1 + \text{cos } \theta)^2}{\text{sin}^2 \theta} \)

Substitute this into \( \frac{p^2 - 1}{p^2 + 1} \):

\( \frac{p^2 - 1}{p^2 + 1} = \frac{\frac{(1 + \text{cos } \theta)^2}{\text{sin}^2 \theta} - 1}{\frac{(1 + \text{cos } \theta)^2}{\text{sin}^2 \theta} + 1} \)

\( = \frac{\frac{(1 + \text{cos } \theta)^2 - \text{sin}^2 \theta}{\text{sin}^2 \theta}}{\frac{(1 + \text{cos } \theta)^2 + \text{sin}^2 \theta}{\text{sin}^2 \theta}} \)

Cancel \( \text{sin}^2 \theta \) from numerator and denominator:

\( = \frac{(1 + \text{cos } \theta)^2 - \text{sin}^2 \theta}{(1 + \text{cos } \theta)^2 + \text{sin}^2 \theta} \)

Expand the terms:

\( = \frac{1 + 2\text{cos } \theta + \text{cos}^2 \theta - \text{sin}^2 \theta}{1 + 2\text{cos } \theta + \text{cos}^2 \theta + \text{sin}^2 \theta} \)

Using \( \text{sin}^2 \theta + \text{cos}^2 \theta = 1 \) and \( \text{sin}^2 \theta = 1 - \text{cos}^2 \theta \):

\( = \frac{1 + 2\text{cos } \theta + \text{cos}^2 \theta - (1 - \text{cos}^2 \theta)}{1 + 2\text{cos } \theta + 1} \)

\( = \frac{1 + 2\text{cos } \theta + \text{cos}^2 \theta - 1 + \text{cos}^2 \theta}{2 + 2\text{cos } \theta} \)

\( = \frac{2\text{cos } \theta + 2\text{cos}^2 \theta}{2(1 + \text{cos } \theta)} \)

Factor out \( 2\text{cos } \theta \) from the numerator:

\( = \frac{2\text{cos } \theta (1 + \text{cos } \theta)}{2(1 + \text{cos } \theta)} \)

Assuming \( 1 + \text{cos } \theta \neq 0 \), we can cancel the term \( 2(1 + \text{cos } \theta) \):

\( = \text{cos } \theta \)

Both methods lead to the same result, confirming that the value of \( \frac{p^2 - 1}{p^2 + 1} \) is indeed \( \text{cos } \theta \).