Problem 187 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}=\frac{\sqrt{3}}{2} $$

Step-by-Step Solution

Answer

The given problem asserts the equality \( \frac{1-\tan ^{2} 15^{\circ}}{1+\tan ^{2} 15^{\circ}}=\frac{\sqrt{3}}{2} \). Proving this involves trigonometric identities and algebraic simplification, which does indeed verify the given equality.

Step 1: Transform the equation

We transform the equation given in the problem using the Pythagorean identity \( \tan^2 \theta = \sec^2 \theta - 1 \). That gives us: \( \frac{1-(\sec^2 15^{\circ}-1)}{1+(\sec^2 15^{\circ}-1)} \)

Step 2: Simplification

Simplifying the equation, we get \( \frac{2-\sec^2 15^{\circ}}{\sec^2 15^{\circ}} \). Further simplifying and writing \( \sec \) in terms of \( \cos \), we get \( \frac{2-\frac{1}{\cos^2 15^{\circ}}}{\frac{1}{\cos^2 15^{\circ}}} = \frac{2\cos^2 15^{\circ}-1}{1} \)

Step 3: Introducing the Double Angle Formula

The expression \(2\cos^2 \theta - 1\) represents the double angle formula for cosine. So we can write: \( \frac{2\cos^2 15^{\circ}-1}{1} = \cos 30^{\circ} \)

Step 4: Evaluating the Cosine Value

We know from the basic values of cosine that \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)

Key Concepts

Tangent and Secant Relationships

Understanding the relationship between the trigonometric functions tangent (\( \tan \)) and secant (\( \sec \)) is vital when dealing with trigonometry problems. The secant function is the reciprocal of the cosine function (\( \sec \theta = \frac{1}{\cos \theta} \)), and the tangent function is the ratio of the sine to the cosine function (\( \tan \theta = \frac{\sin \theta}{\cos \theta} \)).

These relationships are connected by the Pythagorean identity which states that for any angle \( \theta \), \( \tan^2 \theta + 1 = \sec^2 \theta \). This identity can also be rearranged to show that \( \tan^2 \theta = \sec^2 \theta - 1 \), which is particularly useful when simplifying expressions involving tangent and secant functions, as seen in the original exercise.

Trigonometric Simplification

The process of trigonometric simplification involves manipulating trigonometric expressions to a more manageable form. This often includes using fundamental identities and formulas to combine, reduce, or eliminate trigonometric functions.

For example, in the exercise provided, transforming the equation using the Pythagorean identity allowed for an initial simplification. Subsequently, expressing secant in terms of its cosine reciprocal further simplified the expression. This form of simplification is essential when solving trigonometric equations as it can reveal underlying relationships or simpler forms that are easier to evaluate, such as basic trigonometric values or other identities like the double angle formula.

Double Angle Formula

The double angle formulas are a set of three identities that allow the simplification of trigonometric functions of twice a given angle in terms of the functions of the original angle. They are particularly useful in trigonometric simplification and solving equations. The formula for cosine is given by \( \cos(2\theta) = 2\cos^2\theta - 1 \) or \( \cos(2\theta) = 1 - 2\sin^2\theta \).

In the provided exercise, recognizing that \( 2\cos^2 15^\circ - 1 \) corresponds to \( \cos 30^\circ \) via the double angle formula, significantly simplifies the equation and leads to the final evaluation of the trigonometric expression. Understanding and applying the double angle formula can often be the key step in solving more complex trigonometric equations.