Problem 96 - Algebra and Trigonometry

Question

Verify the identity: $$\frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}=4 \tan x \sec x$$

Step-by-Step Solution

Answer

The original equation \(\frac{1+\sin x}{1-\sin x}-\frac{1-\sin x}{1+\sin x}=4 \tan x \sec x\) is indeed an identity. So, the identity has been verified.

Step 1: Transform the left side

Transform both fractions on the left side to have the same denominator. To do this, you multiply the numerators and denominators of both fractions giving the result: \[ \frac{(1 + \sin x)^2 - (1 - \sin x)^2}{(1 - \sin x)(1 + \sin x)}\]

Step 2: Simplify the numerator

Using the difference of squares pattern \(a^2 - b^2 = (a - b)(a + b)\), you can simplify the numerator:\[ \frac{4 \sin x }{(1 - \sin^2 x)}\]

Step 3: Use the Pythagorean Identity

The expression at the denominator \(1 - \sin^2 x\) is equivalent to \(\cos^2 x\) by the Pythagorean identity \( \cos^2 x + \sin^2 x = 1\). So the left side now simplifies to:\[ \frac{4 \sin x }{\cos^2 x}\]

Step 4: Simplify the fraction

Using the properties of fraction division, the left side simplifies even further:\[ 4 \tan x \sec x\]

Key Concepts

Pythagorean identity

In trigonometry, the Pythagorean identity is a fundamental relation that connects the squares of sine and cosine of an angle. The formula is: \[\cos^2 x + \sin^2 x = 1\] This identity is highly useful in simplifying complex trigonometric expressions. In this exercise, it helped transform the original expression by substituting \(1 - \sin^2 x\) with \(\cos^2 x\).

Why is this important? Consider a right triangle. The Pythagorean theorem states \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse. In the unit circle, where the hypotenuse always equals 1, \(\cos^2 x\) and \(\sin^2 x\) become the opposite and adjacent sides squared, explaining how they add up to 1.

For trigonometric identities, this relationship is crucial because:

• It helps to rewrite expressions in simpler terms.
• It allows the transition from sine and cosine terms to tangent or secant and vice versa.
• It forms the basis of many more complex trigonometric identities.

Understanding this identity helps master many trigonometric problems with ease.

difference of squares

The difference of squares is an algebraic pattern which states that any expression of the form \(a^2 - b^2\) can be factored into \((a - b)(a + b)\). This method is pivotal in various mathematical simplifications.

In this exercise, you applied the difference of squares pattern to simplify the numerator of the expression: \[(1 + \sin x)^2 - (1 - \sin x)^2\]

To break it down:

• Recognize \(a = 1 + \sin x\) and \(b = 1 - \sin x\).
• Apply the formula: \((a)^2 - (b)^2 = (a-b)(a+b) = (1 + \sin x - (1 - \sin x))(1 + \sin x + 1 - \sin x)\).
• Once simplified, this results in the expression \(4 \sin x\).

Knowing this pattern can drastically reduce the complexity of polynomial equations and trigonometric expressions, allowing for more straightforward evaluation and a deeper understanding of algebra and its application in trigonometry.

trigonometric simplification

Trigonometric simplification involves breaking down complex expressions into simpler forms using known identities and algebraic techniques. In the given problem, after meeting several algebraic transformations, the equation was transformed to match the right side of the identity.

This process is largely about:

• Identifying trigonometric identities appropriate for substitution.
• Applying algebraic techniques such as factoring, expanding, or combining like terms.
• Utilizing identities like Pythagorean, sum and difference identities, and angle transformations to simplify expressions.

For instance, from :\[\frac{4 \sin x }{\cos^2 x}\]Through knowing the identity \( \tan x = \frac{\sin x}{\cos x} \) and \( \sec x = \frac{1}{\cos x} \), you reorganize the expression to: \[4 \tan x \sec x\]The key is recognizing opportunities to replace complex terms with these identities, achieving the simplest form possible. This results not only in verifying identities but also in solving equations quickly and efficiently.