Question

Which one of the following pairs is correctly matched ? If then (a) \(x=\frac{1+\sin 60^{\circ}-\cos 60^{\circ}}{1+\sin 60^{\circ}+\cos 60^{\circ}} \quad x=\tan 60^{\circ}\) (b) \(x=\frac{1+\sin 90^{\circ}-\cos 90^{\circ}}{1+\sin 90^{\circ}-\cos 90^{\circ}} \quad x=\tan 30^{\circ}\) (c) \(x=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) \(x=\tan 60^{\circ}\) (d) \(x=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}} \quad x=\cos 60^{\circ}\)

Short answer

a) \(x=\frac{1+\sin 60^{\circ}-\cos 60^{\circ}}{1+\sin 60^{\circ}+\cos 60^{\circ}}\) \(x=\tan 60^{\circ}\) (Not a correct pair) b) \(x=\frac{1+\sin 90^{\circ}-\cos 90^{\circ}}{1+\sin 90^{\circ}-\cos 90^{\circ}}\) \(x=\tan 30^{\circ}\) (Not a correct pair) c) \(x=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) \(x=\tan 60^{\circ}\) (A correct pair) d) \(x=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) \(x=\cos 60^{\circ}\) (A correct pair)

Step-by-step solution

Calculate the values for each expression in the pairs

First, we should know the values of trigonometric functions for these angles: \(\sin 30^{\circ} = \frac{1}{2}, \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \tan 30^{\circ} = \frac{1}{\sqrt{3}}\) \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}, \cos 60^{\circ} = \frac{1}{2}, \tan 60^{\circ} = \sqrt{3}\) \(\sin 90^{\circ} = 1, \cos 90^{\circ} = 0, \tan 90^{\circ} = \text{undefined}\) Now, let's calculate the values of the expressions on both sides of the pairs.

Pair (a): \(x=\frac{1+\sin 60^{\circ}-\cos 60^{\circ}}{1+\sin 60^{\circ}+\cos 60^{\circ}} \quad x=\tan 60^{\circ}\)

The left side expression: \(x = \frac{1+\frac{\sqrt{3}}{2}-\frac{1}{2}}{1+\frac{\sqrt{3}}{2}+\frac{1}{2}} = 1\) The right side expression: \(x = \tan 60^{\circ} = \sqrt{3}\) The expressions don't match, not a correct pair.

Pair (b): \(x=\frac{1+\sin 90^{\circ}-\cos 90^{\circ}}{1+\sin 90^{\circ}-\cos 90^{\circ}} \quad x=\tan 30^{\circ}\)

The left side expression: \(x = \frac{1+1-0}{1+1-0} = 1\) The right side expression: \(x = \tan 30^{\circ} = \frac{1}{\sqrt{3}}\) The expressions don't match, not a correct pair.

Pair (c): \(x=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}} \quad x=\tan 60^{\circ}\)

The left side expression: \(x = \frac{2\times \frac{1}{\sqrt{3}}}{1-(\frac{1}{\sqrt{3}})^{2}} = \frac{2}{\sqrt{3}}\times \frac{\frac{2}{3}}{1-\frac{1}{3}} = \frac{4}{3\sqrt{3}}\times \frac{2}{3} = \sqrt{3}\) The right side expression: \(x = \tan 60^{\circ} = \sqrt{3}\) The expressions match, a correct pair.

Pair (d): \(x=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}} \quad x=\cos 60^{\circ}\)

The left side expression: \(x = \frac{1-(\frac{1}{\sqrt{3}})^{2}}{1+(\frac{1}{\sqrt{3}})^{2}} = \frac{1-\frac{1}{3}}{1+\frac{1}{3}} = \frac{\frac{2}{3}}{\frac{4}{3}} = \frac{1}{2}\) The right side expression: \(x = \cos 60^{\circ} = \frac{1}{2}\) The expressions match, a correct pair. So, the correct pairs are: (c) \(x=\frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}\) \(x=\tan 60^{\circ}\) (d) \(x=\frac{1-\tan ^{2} 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\) \(x=\cos 60^{\circ}\)

Key concepts

Sine and Cosine Values

Understanding the values of sine and cosine functions is pivotal in mastering trigonometry. The sine function relates to the ratio of the length of the side of the triangle opposite the angle to the length of the hypotenuse. In symbol form, if \( \theta \) is one of the non-right angles in a right triangle, then \( \text{sin}(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}} \). Similarly, the cosine function gives the ratio of the length of the adjacent side to the hypotenuse. That is, \( \text{cos}(\theta) = \frac{\text{adjacent side}}{\text{hypotenuse}} \).

For common angles in trigonometry, such as \( 30^\text{o} \), \( 60^\text{o} \), and \( 90^\text{o} \), the sine and cosine values are derived from the geometrical properties of an equilateral triangle and the unit circle. For instance, \( \text{sin}(60^\text{o}) = \frac{\text{opposite side}}{\text{hypotenuse}} = \frac{\text{height of equilateral triangle}}{\text{length of side}} = \frac{\frac{\text{sqrt}(3)}{2}}{1} = \frac{\text{sqrt}(3)}{2} \) and \( \text{cos}(60^\text{o}) = \frac{1}{2} \). Remembering these values is crucial when solving problems involving trigonometric functions.

Tangent Function

The tangent function, often abbreviated as \( \text{tan} \), is another fundamental trigonometric ratio. It is defined as the ratio of the sine value to the cosine value for a particular angle, or simply, the ratio of the opposite side to the adjacent side in a right-angled triangle. Thus, \( \text{tan}(\theta) = \frac{\text{sin}(\theta)}{\text{cos}(\theta)}\) or \( \text{tan}(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} \).

For example, the \( \text{tan}(30^\text{o}) \) can be computed using the known sine and cosine values for \( 30^\text{o} \), giving us \( \text{tan}(30^\text{o}) = \frac{\text{sin}(30^\text{o})}{\text{cos}(30^\text{o})} = \frac{\frac{1}{2}}{\frac{\text{sqrt}(3)}{2}} = \frac{1}{\text{sqrt}(3)} \). Understanding how the tangent function relates to sine and cosine is essential for solving trigonometric equations and verifying identities.

Trigonometric Ratio Calculations

Trigonometric ratio calculations form the essence of solving complex trigonometric problems. These include operations like solving for an unknown side or angle in a right triangle, verifying trigonometric identities, and simplifying trigonometric expressions.

For successful calculations, recognizing patterns and applying trigonometric identities such as the Pythagorean identity \( \text{sin}^2(\theta) + \text{cos}^2(\theta) = 1 \), and the angle sum and difference formulas can be very helpful. Trigonometry also makes use of reciprocal identities, like \( \text{cot}(\theta) = \frac{1}{\text{tan}(\theta)} \), ratio identities, and even more complex relations like those involving the double angle or half angle.

As seen in the exercise example, when given expressions such as \( \frac{2 \text{tan}(30^\text{o})}{1-\text{tan}^2(30^\text{o})} \), we apply the value of \( \text{tan}(30^\text{o}) \) and simplify using algebraic techniques to find that it equals \( \text{tan}(60^\text{o}) \). This process of substituting known trigonometric values and applying identities is fundamental to verifying trigonometric equations.