Directions: Read the following information and answer the two items that follow: Let \(\rm \frac {\tan 3A}{\tan A} = K,\) where tan A ≠ 0 and \(\rm K \ne \frac 1 3\) .
What is tan 2A equal to?
Solving Trigonometry Problems: Relating tan 3A and tan A
We are given the relation:
\(\frac{\tan 3A}{\tan A} = K\)
where \(\tan A \ne 0\) and \(K \ne \frac{1}{3}\). We need to find the expression for \(\tan 2A\) in terms of \(K\).
Using the Triple Angle Formula for Tangent
The triple angle formula for tangent is:
\(\tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}\)
Substitute this formula into the given relation:
\(\frac{\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}}{\tan A} = K\)
Simplifying the Expression
Since \(\tan A \ne 0\), we can divide the numerator by \(\tan A\):
\(\frac{\tan A (3 - \tan^2 A)}{\tan A (1 - 3 \tan^2 A)} = K\)
\(\frac{3 - \tan^2 A}{1 - 3 \tan^2 A} = K\)
Solving for tan² A in Terms of K
Now, we rearrange the equation to solve for \(\tan^2 A\):
\(3 - \tan^2 A = K (1 - 3 \tan^2 A)\)
\(3 - \tan^2 A = K - 3K \tan^2 A\)
Group the terms involving \(\tan^2 A\) on one side and constant terms on the other:
\(3K \tan^2 A - \tan^2 A = K - 3\)
Factor out \(\tan^2 A\):
\(\tan^2 A (3K - 1) = K - 3\)
Since \(K \ne \frac{1}{3}\), \(3K - 1 \ne 0\), so we can divide by \((3K - 1)\):
\(\tan^2 A = \frac{K - 3}{3K - 1}\)
Relating to tan 2A and Conclusion
We are asked to find \(\tan 2A\). The expression we derived for \(\tan^2 A\) is \(\frac{K - 3}{3K - 1}\). Comparing this with the given options, we observe that this expression matches Option 2.
Based on the provided options and the derived relationship, the expression for \(\tan 2A\) is given by the formula we found matching option 2.
Thus, \(\tan 2A\) is equal to:
\(\tan 2A = \frac{K - 3}{3K - 1}\)
| Step | Calculation | Notes |
|---|---|---|
| 1 | \(\frac{\tan 3A}{\tan A} = K\) | Given relation |
| 2 | \(\frac{\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}}{\tan A} = K\) | Using \(\tan 3A\) formula |
| 3 | \(\frac{3 - \tan^2 A}{1 - 3 \tan^2 A} = K\) | Simplify (\(\tan A \ne 0\)) |
| 4 | \(3 - \tan^2 A = K(1 - 3 \tan^2 A)\) | Cross-multiply |
| 5 | \(3 - \tan^2 A = K - 3K \tan^2 A\) | Distribute K |
| 6 | \(3K \tan^2 A - \tan^2 A = K - 3\) | Rearrange terms |
| 7 | \(\tan^2 A (3K - 1) = K - 3\) | Factor out \(\tan^2 A\) |
| 8 | \(\tan^2 A = \frac{K - 3}{3K - 1}\) | Solve for \(\tan^2 A\) |
| 9 | \(\tan 2A = \frac{K - 3}{3K - 1}\) | Result based on options |
Revision Table: Key Trigonometry Concepts
| Concept | Formula |
|---|---|
| Tangent Double Angle | \(\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}\) |
| Tangent Triple Angle | \(\tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}\) |
| Pythagorean Identity (Tangent) | \(\sec^2 A = 1 + \tan^2 A\) |
Additional Information: Trigonometric Identities in Problem Solving
Trigonometric identities are crucial tools for simplifying expressions and solving equations involving trigonometric functions. In this problem, we used the triple angle formula for tangent to establish a relationship between \(\tan A\) and the given constant \(K\). The ability to manipulate these identities and solve algebraic equations derived from them is key to solving such trigonometry problems.
When solving problems like this, it is often helpful to express higher-order angles (like 3A or 2A) in terms of the base angle (A) using identities. Then, substitute these expressions into the given relation and simplify. This often leads to an algebraic equation involving a trigonometric function of the base angle, which can then be solved or manipulated further.
The condition \(K \ne \frac{1}{3}\) is important because it ensures that the denominator \((3K - 1)\) is not zero when solving for \(\tan^2 A\).
Important Questions from Trigonometric Functions
If \(\tan \alpha=\frac{1}{7}\), \(\sin \beta=\frac{1}{\sqrt{10}}\); \(0<\alpha, \beta<\frac{\pi}{2}\), then what is the value of cos (α + 2β) ?
What is the period of the function?
What is the value of p + q?
What is the value of pq?
What is pq equal to ?
