Understanding the Trigonometry Problem

The question provides a relationship between the ratio of \(\sin(x+y)\) and \(\sin(x-y)\) and the ratio of two variables, \(a\) and \(b\). Specifically, it is given that:

We are asked to find the value of the ratio \(\frac{{\tan x}}{{\tan y}}\).

To solve this problem, we will use trigonometric identities and a property of ratios called Componendo and Dividendo.

Applying Componendo and Dividendo

The property of Componendo and Dividendo states that if \(\frac{A}{B} = \frac{C}{D}\), then \(\frac{A + B}{A - B} = \frac{C + D}{C - D}\). This property is very useful when dealing with ratios like the one given in the question.

Let's apply this property to the given equation:

Left side: \(A = \sin(x+y)\), \(B = \sin(x-y)\)

Right side: \(C = a+b\), \(D = a-b\)

Applying Componendo and Dividendo:

Simplifying the Right Side

Let's simplify the right side of the equation first:

So the equation becomes:

Using Sum-to-Product Identities

Now, let's simplify the left side of the equation using the sum-to-product trigonometric identities:

• \(\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)\)
• \(\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)\)

In our case, \(A = x+y\) and \(B = x-y\).

• \(A+B = (x+y) + (x-y) = 2x\)
• \(A-B = (x+y) - (x-y) = 2y\)

Applying the identities to the left side:

Numerator: \(\sin(x+y) + \sin(x-y) = 2 \sin\left(\frac{2x}{2}\right) \cos\left(\frac{2y}{2}\right) = 2 \sin x \cos y\)

Denominator: \(\sin(x+y) - \sin(x-y) = 2 \cos\left(\frac{2x}{2}\right) \sin\left(\frac{2y}{2}\right) = 2 \cos x \sin y\)

So the left side simplifies to:

Equating and Finding tan x / tan y

Now we equate the simplified left side and the simplified right side:

We can rewrite the left side by separating the terms involving x and y:

Using the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), this becomes:

Since \(\cot y = \frac{1}{{\tan y}}\), we can substitute this into the equation:

Thus, the ratio \(\frac{{\tan x}}{{\tan y}}\) is equal to \(\frac{a}{b}\).

Final Answer Check

We started with the given equation \(\frac{{\sin \left( {x\; + \;y} \right)}}{{\sin \left( {x\; - \;y} \right)}} = \;\frac{{a\; + \;b}}{{a\; - \;b}}\). By applying Componendo and Dividendo and using sum-to-product identities, we successfully transformed the equation to find the ratio \(\frac{{\tan x}}{{\tan y}}\) in terms of \(a\) and \(b\). The result obtained is \(\frac{a}{b}\), which matches one of the given options.

Revision Table: Key Concepts Review

Additional Information: Ratio and Proportion Basics

The concept of ratio and proportion, especially the Componendo and Dividendo rule, is fundamental in solving many algebraic and trigonometric problems involving fractions or ratios. Understanding how to manipulate equations using these properties can significantly simplify complex expressions.

Componendo and Dividendo is derived from basic algebraic manipulation:

1. Start with \(\frac{A}{B} = \frac{C}{D}\).
2. Add 1 to both sides: \(\frac{A}{B} + 1 = \frac{C}{D} + 1 \implies \frac{A+B}{B} = \frac{C+D}{D}\) (Componendo).
3. Subtract 1 from both sides: \(\frac{A}{B} - 1 = \frac{C}{D} - 1 \implies \frac{A-B}{B} = \frac{C-D}{D}\) (Dividendo).
4. Divide the Componendo result by the Dividendo result: \(\frac{\frac{A+B}{B}}{\frac{A-B}{B}} = \frac{\frac{C+D}{D}}{\frac{C-D}{D}} \implies \frac{A+B}{A-B} = \frac{C+D}{C-D}\).

This method is particularly effective when you have a ratio equal to another ratio, like in this trigonometry problem involving \(\sin(x+y)/\sin(x-y)\).