Problem 153 - Calculus Volume 1

Question

For the following exercises, verify that each equation is an identity. $$\frac{1}{1-\sin \alpha}+\frac{1}{1+\sin \alpha}=2 \sec ^{2} \alpha$$

Step-by-Step Solution

Answer

The equation is an identity as both sides simplify to \(2 \sec^2 \alpha\).

Step 1: Common Denominator

Combine the fractions on the left-hand side of the equation by finding a common denominator. The common denominator for \( \frac{1}{1-\sin \alpha} \) and \( \frac{1}{1+\sin \alpha} \) is \((1-\sin \alpha)(1+\sin \alpha)\). Thus,\[\frac{1}{1-\sin \alpha} + \frac{1}{1+\sin \alpha} = \frac{(1+\sin \alpha) + (1-\sin \alpha)}{(1-\sin \alpha)(1+\sin \alpha)}.\]Simplify the numerator to \(2\).

Step 2: Simplify Denominator

The denominator \((1-\sin \alpha)(1+\sin \alpha)\) can be simplified using the difference of squares formula:\[ (1-\sin \alpha)(1+\sin \alpha) = 1 - (\sin^2 \alpha).\]

Step 3: Use Pythagorean Identity

Recall the Pythagorean identity: \( \sin^2 \alpha + \cos^2 \alpha = 1 \). Thus, we can replace \(1 - \sin^2 \alpha\) with \( \cos^2 \alpha\). This gives us: \[\frac{2}{1-\sin^2 \alpha} = \frac{2}{\cos^2 \alpha}.\]

Step 4: Use Secant Identity

The reciprocal identity states that \( \frac{1}{\cos^2 \alpha} = \sec^2 \alpha\). Therefore, \[\frac{2}{\cos^2 \alpha} = 2 \sec^2 \alpha.\]Thus, the right-hand side of the original equation \(2 \sec^2 \alpha\) is equal to the simplified left-hand side.

Key Concepts

Common Denominator

Combining fractions is an important skill in algebra and trigonometry. When fractions have different denominators, you must find a common denominator to combine them. In the exercise, the fractions are \( \frac{1}{1-\sin \alpha} \) and \( \frac{1}{1+\sin \alpha} \).
To combine these, you must find a denominator that both terms can share, which simplifies calculations. The common denominator is \((1-\sin \alpha)(1+\sin \alpha)\).

• Rewrite each fraction using this common denominator.
• This allows the numerators to simply add up.

For these particular fractions, the numerator of the combined fraction becomes \((1+\sin \alpha) + (1-\sin \alpha)\), which simplifies exactly to \(2\).
This is the streamlined approach to dealing with fractions in equations.

Difference of Squares

The difference of squares is a concept that can simplify expressions within mathematical equations. It refers to an expression of the form \((a-b)(a+b)\), and its result is \(a^2 - b^2\). In this context, \((1-\sin \alpha)(1+\sin \alpha)\) is perfectly structured to apply this principle because it matches the pattern.

• Identify \(a = 1\) and \(b = \sin \alpha\).
• Applying the formula gives \(1 - \sin^2 \alpha\).

This identity is crucial for simplifying the fraction's denominator, which will then allow for further simplification using another trigonometric identity. Recognizing when to use difference of squares is essential for reducing complexity in expressions.

Pythagorean Identity

The Pythagorean identity is one of the most fundamental identities in trigonometry. It states that \(\sin^2 \alpha + \cos^2 \alpha = 1\). This identity links the sine and cosine functions, allowing them to be interchanged or simplified in many situations.
In our problem, it lets us replace \(1 - \sin^2 \alpha\) with \(\cos^2 \alpha\) due to the identity:

• \(1 - \sin^2 \alpha = \cos^2 \alpha\)

With this substitution, the equation becomes much simpler, turning \(\frac{2}{1-\sin^2 \alpha}\) into \(\frac{2}{\cos^2 \alpha}\). This substitution is often a gateway for further simplifications and transformations in equations.

Secant Function

The secant function is closely related to the cosine function. It is the reciprocal, which means \(\sec \alpha = \frac{1}{\cos \alpha}\). By understanding the secant function, you can interchangeably use secant and cosine in expressions by applying this relationship.
For the given problem, after applying the Pythagorean identity, you end up with the expression \(\frac{2}{\cos^2 \alpha}\).

• Recognize that \(\frac{1}{\cos^2 \alpha}\) is simply \(\sec^2 \alpha\).
• This transforms the equation to \(2 \sec^2 \alpha\).

This final step completes the verification of the identity, showcasing how the secant function acts as a bridge in converting between different trigonometric functions efficiently.