If 3 tanθ = \(2\sqrt 3 \) sinθ, 0° < θ < 90°, then the value of \(\rm \frac{{\cos e{c^2}2\,\theta + {{\cot }^2}2\,\theta }}{{{{\sin }^2}\,\theta + {{\tan }^2}2\,\theta }}\) is:
Solving Trigonometric Equations and Evaluating Expressions
The problem requires us to first solve a given trigonometric equation to find the value of the angle θ within a specified range, and then substitute this value of θ into a complex trigonometric expression to find its numerical value.
The given equation is: \(3 \tan \theta = 2\sqrt 3 \) \(\sin \theta\), with the condition \(0^\circ < \theta < 90^\circ\).
Step-by-step Solution to Find θ
We start by rewriting the tangent function in terms of sine and cosine:
\[ 3 \frac{\sin \theta}{\cos \theta} = 2\sqrt 3 \sin \theta \] Since \(0^\circ < \theta < 90^\circ\), the value of \(\sin \theta\) is not zero. Therefore, we can safely divide both sides of the equation by \(\sin \theta\):
\[ \frac{3}{\cos \theta} = 2\sqrt 3 \] Now, we can solve for \(\cos \theta\):
\[ \cos \theta = \frac{3}{2\sqrt 3} \] To simplify the right side, we can multiply the numerator and denominator by \(\sqrt 3\):
\[ \cos \theta = \frac{3 \times \sqrt 3}{2\sqrt 3 \times \sqrt 3} = \frac{3\sqrt 3}{2 \times 3} = \frac{\sqrt 3}{2} \] We need to find the value of θ in the range \(0^\circ < \theta < 90^\circ\) for which \(\cos \theta = \frac{\sqrt 3}{2}\). The angle in this range satisfying this condition is \(30^\circ\).
Thus, \(\theta = 30^\circ\).
Evaluating the Trigonometric Expression
Now that we have found \(\theta = 30^\circ\), we need to evaluate the expression: \(\rm \frac{{\cos e{c^2}2\,\theta + {{\cot }^2}2\,\theta }}{{{{\sin }^2}\,\theta + {{\tan }^2}2\,\theta }}\).
First, let's find the values of the trigonometric functions for \(\theta = 30^\circ\) and \(2\theta = 60^\circ\).
- \(\sin \theta = \sin 30^\circ = \frac{1}{2}\)
- \(\sin^2 \theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\)
- \(2\theta = 2 \times 30^\circ = 60^\circ\)
- \(\tan 2\theta = \tan 60^\circ = \sqrt 3\)
- \(\tan^2 2\theta = (\sqrt 3)^2 = 3\)
- \(\sin 2\theta = \sin 60^\circ = \frac{\sqrt 3}{2}\)
- \(\cos ec 2\theta = \frac{1}{\sin 2\theta} = \frac{1}{\sqrt 3 / 2} = \frac{2}{\sqrt 3}\)
- \(\cos ec^2 2\theta = \left(\frac{2}{\sqrt 3}\right)^2 = \frac{4}{3}\)
- \(\cot 2\theta = \frac{1}{\tan 2\theta} = \frac{1}{\sqrt 3}\)
- \(\cot^2 2\theta = \left(\frac{1}{\sqrt 3}\right)^2 = \frac{1}{3}\)
Now, substitute these values into the given expression:
Numerator: \(\cos ec^2 2\,\theta + \cot^2 2\,\theta = \frac{4}{3} + \frac{1}{3} = \frac{4+1}{3} = \frac{5}{3}\)
Denominator: \(\sin^2 \,\theta + \tan^2 2\,\theta = \frac{1}{4} + 3 = \frac{1}{4} + \frac{12}{4} = \frac{1+12}{4} = \frac{13}{4}\)
The expression is \(\frac{\text{Numerator}}{\text{Denominator}}\):
\[ \frac{\frac{5}{3}}{\frac{13}{4}} = \frac{5}{3} \times \frac{4}{13} = \frac{5 \times 4}{3 \times 13} = \frac{20}{39} \] The value of the expression is \(\frac{20}{39}\).
| Trigonometric Function | Value at \(\theta = 30^\circ\) | Value at \(2\theta = 60^\circ\) | Squared Value at \(\theta = 30^\circ\) | Squared Value at \(2\theta = 60^\circ\) |
|---|---|---|---|---|
| \(\sin\) | \(\frac{1}{2}\) | \(\frac{\sqrt 3}{2}\) | \(\frac{1}{4}\) (\(\sin^2 \theta\)) | - |
| \(\tan\) | \(\frac{1}{\sqrt 3}\) | \(\sqrt 3\) | - | 3 (\(\tan^2 2\theta\)) |
| \(\cos ec\) | 2 | \(\frac{2}{\sqrt 3}\) | - | \(\frac{4}{3}\) (\(\cos ec^2 2\theta\)) |
| \(\cot\) | \(\sqrt 3\) | \(\frac{1}{\sqrt 3}\) | - | \(\frac{1}{3}\) (\(\cot^2 2\theta\)) |
Using these squared values in the expression:
Numerator: \(\frac{4}{3} + \frac{1}{3} = \frac{5}{3}\)
Denominator: \(\frac{1}{4} + 3 = \frac{1}{4} + \frac{12}{4} = \frac{13}{4}\)
Result: \(\frac{5/3}{13/4} = \frac{5}{3} \times \frac{4}{13} = \frac{20}{39}\)
Conclusion
By solving the trigonometric equation \(3 \tan \theta = 2\sqrt 3 \sin \theta\), we found that \(\theta = 30^\circ\) for the given range. Substituting this value into the expression \(\rm \frac{{\cos e{c^2}2\,\theta + {{\cot }^2}2\,\theta }}{{{{\sin }^2}\,\theta + {{\tan }^2}2\,\theta }}\) yields the value \(\frac{20}{39}\).
Revision Table: Key Trigonometric Values
| Angle | \(\sin\) | \(\cos\) | \(\tan\) | \(\cos ec\) | \(\cot\) |
|---|---|---|---|---|---|
| \(30^\circ\) | \(\frac{1}{2}\) | \(\frac{\sqrt 3}{2}\) | \(\frac{1}{\sqrt 3}\) | 2 | \(\sqrt 3\) |
| \(60^\circ\) | \(\frac{\sqrt 3}{2}\) | \(\frac{1}{2}\) | \(\sqrt 3\) | \(\frac{2}{\sqrt 3}\) | \(\frac{1}{\sqrt 3}\) |
Additional Information: Trigonometric Identities and Definitions
Solving this problem involved using fundamental trigonometric definitions and values for standard angles. Key points include:
- The definition of tangent: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). This was crucial in simplifying the initial equation.
- Reciprocal identities: \(\cos ec \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{1}{\tan \theta}\). These were used to calculate the values of \(\cos ec 2\theta\) and \(\cot 2\theta\).
- Understanding the values of trigonometric functions for standard angles like \(30^\circ\) and \(60^\circ\) is essential for solving many trigonometry problems quickly and accurately.
- The domain \(0^\circ < \theta < 90^\circ\) is important because it tells us that \(\sin \theta > 0\), \(\cos \theta > 0\), and \(\tan \theta > 0\), and that the angle is in the first quadrant. This helps confirm that dividing by \(\sin \theta\) is valid and that the value of \(\theta\) we found is the unique solution in this range.
Practice with trigonometric equations and evaluating expressions builds confidence for exam situations.
Important Questions from Trigonometry
The given equation can be reduced to
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