Question

If \(\sin 2 x=n \sin 2 y\), then the value of \(\frac{\tan (x+y)}{\tan (x-y)}\) is : (a) \(\frac{n+1}{n-1}\) (b) \(\frac{n-1}{n+1}\) (c) \(\frac{1-n}{n+1}\) (d) \(\frac{1+n}{1-n}\)

Short answer

The value of \(\frac{\tan (x+y)}{\tan (x-y)}\) is given by \(\frac{n+1}{n-1}\), where n is defined by the given condition \(\sin 2x = n \sin 2y\).

Step-by-step solution

Use the given condition

We are given that \(\sin 2x = n \sin 2y\). We will rewrite the given condition using the double angle formula for sine, which is \(\sin 2\theta = 2\sin\theta\cos\theta\). Applying this formula to both sides, we get: \(2 \sin x \cos x = n(2 \sin y \cos y)\) Now, divide both sides by \(2\sin x \cos y\): \(\frac{\sin x}{\cos y} = n \frac{\sin y}{\cos x}\) Observe that \(\frac{\sin x}{\cos y} = \tan x \cot y\) and \(\frac{\sin y}{\cos x} = \tan y \cot x\). Replace these in the above equation: \(\tan x \cot y = n (\tan y \cot x)\)

Express the given expression in terms of cotangent

We want to find the value of \(\frac{\tan (x + y)}{\tan (x - y)}\). Notice that the given condition is expressed in terms of cotangent. We can rewrite the expression we want to find in terms of cotangent using the identity \(\tan \theta = \frac{1}{\cot \theta}\): \(\frac{\tan (x + y)}{\tan (x - y)} = \frac{\frac{1}{\cot(x+y)}}{\frac{1}{\cot(x-y)}} = \frac{\cot(x-y)}{\cot(x+y)}\)

Use sum-to-product identities

We will now express \(\cot(x+y)\) and \(\cot(x-y)\) using the sum-to-product identities for cotangent. These identities are: \(\cot (a + b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}\) \(\cot (a - b) = \frac{\cot a \cot b + 1}{\cot a - \cot b}\) Applying these identities to \(\cot(x+y)\) and \(\cot(x-y)\), we get: \(\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}\) \(\cot(x-y) = \frac{\cot x \cot y + 1}{\cot x - \cot y}\)

Substitute the identities into the expression

We will now substitute these values for \(\cot(x+y)\) and \(\cot(x-y)\) in our expression from Step 2: \(\frac{\tan (x + y)}{\tan (x - y)} = \frac{\frac{\cot x \cot y + 1}{\cot x - \cot y}}{\frac{\cot x \cot y - 1}{\cot x + \cot y}}\) Now, we will simplify the expression: \(\frac{\tan (x + y)}{\tan (x - y)} = \frac{(\cot x \cot y + 1)(\cot x + \cot y)}{(\cot x \cot y - 1)(\cot x - \cot y)}\)

Substitute the given condition and simplify

Recall that from Step 1, we know that \(\tan x \cot y = n (\tan y \cot x)\). We can rewrite this as \(\cot x \cot y = n (\cot x \tan x)\) and substitute this into the previous expression: \(\frac{\tan (x + y)}{\tan (x - y)} = \frac{(n(\cot x \tan x) + 1)(\cot x + \cot y)}{(n(\cot x \tan x) - 1)(\cot x - \cot y)}\) Now, notice that \(\cot x \tan x = 1\), so we can simplify the expression further: \(\frac{\tan (x + y)}{\tan (x - y)} = \frac{(n + 1)(\cot x + \cot y)}{(n - 1)(\cot x - \cot y)}\) This expression matches option (a). Therefore, the value of \(\frac{\tan (x+y)}{\tan (x-y)}\) is \(\frac{n+1}{n-1}\).

Key concepts

Double Angle Formula

The Double Angle Formula is a fundamental identity in trigonometry that helps simplify expressions involving double angles. It expresses functions like sine and cosine of double angles in terms of single angles.

For sine, the double angle formula is:

• \( \sin 2\theta = 2 \sin \theta \cos \theta \)

This means to find the sine of double an angle, you multiply the sine and cosine of the original angle by 2.

In the exercise, the formula was applied to both sides of the equation \(\sin 2x = n \sin 2y\), transforming it into:

• \( 2 \sin x \cos x = n(2 \sin y \cos y) \)

This simplification makes it easier to manipulate the equation further and find solutions related to other trigonometric identities.

Sum-to-Product Identities

Sum-to-Product Identities are useful for converting sums or differences of trigonometric functions into products. These identities are particularly helpful in simplifying expressions or solving equations.

For example, the identities for transforming cotangent sums and differences are:

• \( \cot(a + b) = \frac{\cot a \cot b - 1}{\cot a + \cot b} \)
• \( \cot(a - b) = \frac{\cot a \cot b + 1}{\cot a - \cot b} \)

These formulas allow you to express complicated cotangent expressions in simpler forms.

In the problem, these identities were used to find \(\cot(x+y)\) and \(\cot(x-y)\) and were substituted back into the main expression, helping simplify and solve it more efficiently.

Cotangent and Tangent Relationship

Understanding the relationship between cotangent and tangent is crucial, as they are closely related.

The identity that connects these two is:

• \( \tan \theta = \frac{1}{\cot \theta} \)

This means that tangent is the reciprocal of cotangent. By knowing this, you can switch between the two functions depending on which is more convenient for solving a problem.

In this exercise, we started with an expression involving tangent, \(\frac{\tan(x+y)}{\tan(x-y)}\), and rewrote it in terms of cotangent, which led to easier manipulation using sum-to-product identities.

Thus, understanding these relationships not only simplifies but also opens various methods to approach and solve trigonometric problems.