If \(\frac{{\tan \theta + \sin \theta }}{{\tan \theta - \sin \theta }} = \frac{{k + 1}}{{k - 1}},\) then k = ?
sec θ
Solving Trigonometric Equations: Finding k
The problem asks us to find the value of 'k' given a specific trigonometric equation relating \(\tan \theta\) and \(\sin \theta\).
The given equation is:
\[ \frac{{\tan \theta + \sin \theta }}{{\tan \theta - \sin \theta }} = \frac{{k + 1}}{{k - 1}} \]
This equation is in a form that is perfectly suited for applying a useful property from ratios and proportions known as Componendo and Dividendo.
Understanding Componendo and Dividendo
The Componendo and Dividendo rule states that if \( \frac{a}{b} = \frac{c}{d} \), then \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \). Conversely, and relevant to our problem, if \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \), then \( \frac{a}{b} = \frac{c}{d} \).
Applying the Rule to the Trigonometric Equation
Let's compare the given equation to the form \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \):
- On the left side, we have \( a = \tan \theta \) and \( b = \sin \theta \).
- On the right side, we have \( c = k \) and \( d = 1 \).
Applying the converse of the Componendo and Dividendo rule, which states that if \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \), then \( \frac{a}{b} = \frac{c}{d} \), we get:
\[ \frac{\tan \theta}{\sin \theta} = \frac{k}{1} \]
This simplifies to:
\[ k = \frac{\tan \theta}{\sin \theta} \]
Simplifying the Expression for k
Now, we need to simplify the right side of the equation \( k = \frac{\tan \theta}{\sin \theta} \). We know the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Let's substitute this into the expression for k:
\[ k = \frac{\frac{\sin \theta}{\cos \theta}}{\sin \theta} \]
To simplify this complex fraction, we can write it as the numerator multiplied by the reciprocal of the denominator:
\[ k = \frac{\sin \theta}{\cos \theta} \times \frac{1}{\sin \theta} \]
Assuming \(\sin \theta \neq 0\), we can cancel out \(\sin \theta\) from the numerator and the denominator:
\[ k = \frac{1}{\cos \theta} \]
Finally, we know the reciprocal identity \( \frac{1}{\cos \theta} = \sec \theta \).
So, the value of k is:
\[ k = \sec \theta \]
Comparing with Options
Let's compare our result with the given options:
| Option | Value |
|---|---|
| 1 | \( \cos \theta \) |
| 2 | \( \sec \theta \) |
| 3 | \( \sin \theta \) |
| 4 | \( \operatorname{cosec} \theta \) |
Our calculated value for k is \( \sec \theta \), which matches Option 2.
Conclusion
By using the Componendo and Dividendo rule and simplifying the resulting trigonometric expression, we found that k is equal to \( \sec \theta \).
Revision Table: Key Trigonometry Concepts
| Concept | Description | Formula(s) |
|---|---|---|
| Tangent (\(\tan \theta\)) | Ratio of the opposite side to the adjacent side in a right-angled triangle; also ratio of sin to cos. | \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sin \theta}{\cos \theta} \) |
| Sine (\(\sin \theta\)) | Ratio of the opposite side to the hypotenuse in a right-angled triangle. | \( \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \) |
| Secant (\(\sec \theta\)) | Reciprocal of cosine; ratio of the hypotenuse to the adjacent side. | \( \sec \theta = \frac{1}{\cos \theta} = \frac{\text{Hypotenuse}}{\text{Adjacent}} \) |
| Componendo and Dividendo | A rule relating ratios: If \( \frac{a}{b} = \frac{c}{d} \), then \( \frac{a+b}{a-b} = \frac{c+d}{c-d} \) and vice versa. | \( \frac{a}{b} = \frac{c}{d} \iff \frac{a+b}{a-b} = \frac{c+d}{c-d} \) |
Additional Information: Trigonometric Identities and Rules
Solving trigonometric equations often involves using fundamental identities and algebraic rules like Componendo and Dividendo. Here are some important points:
- Reciprocal Identities: These relate the six trigonometric functions to each other. Examples include \( \sec \theta = \frac{1}{\cos \theta} \), \( \operatorname{cosec} \theta = \frac{1}{\sin \theta} \), \( \cot \theta = \frac{1}{\tan \theta} \).
- Quotient Identities: These express tangent and cotangent in terms of sine and cosine. The most common is \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
- Pythagorean Identities: These are derived from the Pythagorean theorem and are crucial for simplifying expressions. The main one is \( \sin^2 \theta + \cos^2 \theta = 1 \). Others include \( 1 + \tan^2 \theta = \sec^2 \theta \) and \( 1 + \cot^2 \theta = \operatorname{cosec}^2 \theta \).
- Algebraic Manipulations: Besides specific trigonometric identities, standard algebraic techniques like factoring, finding common denominators, and applying rules like Componendo and Dividendo are frequently used. Recognizing patterns like \( \frac{a+b}{a-b} \) is key to applying appropriate rules.
Practicing with various trigonometric problems helps in quickly identifying which identities or rules are most effective for simplification and solving.
Important Questions from Trigonometric Ratios and Identities
If cosec θ = 13/12, then sin θ + cos θ - tan θ is equal to:
What is the value of \(\frac{3 \sin 58^{\circ}}{\cos 32^{\circ}}+\frac{3 \sin 42^{\circ}}{\cos 48^{\circ}}\) ?
If α + β = 90° and α = 2β, then the value of 3 cos 2 α - 2 sin 2 β is equal to:
If \(\sqrt{3}\) tan θ = 3 sin θ, then what is the value of sin 2θ − cos 2θ ?
- \(\frac{(1 + tan\theta + sec\theta)(1 + cot\theta - cosec\theta)}{(sec\theta + tan\theta)(1 - sin\theta)}\) is equal to:
