If \(\frac{{\sec \theta \; - \;\tan \theta }}{{\sec \theta \; + \;\tan \theta }}\; = \;\frac{3}{5},\) then the value of \(\frac{{cosec\theta \; + \;\cot \theta }}{{cosec\;\theta \; - {\rm{\;cot}}\theta }}\) is:
31 + 8√15
Understanding the Trigonometric Problem
The problem asks us to find the value of a trigonometric expression involving cosecant and cotangent, given a ratio involving secant and tangent. We are provided with the equation: \(\frac{{\sec \theta \; - \;\tan \theta }}{{\sec \theta \; + \;\tan \theta }}\; = \;\frac{3}{5}\).
Our goal is to find the value of: \(\frac{{cosec\theta \; + \;\cot \theta }}{{cosec\;\theta \; - {\rm{\;cot}}\theta }}\).
Applying Componendo and Dividendo to the Secant and Tangent Ratio
The given equation has the form \(\frac{a-b}{a+b} = \frac{c}{d}\). A useful technique for solving such equations is applying the Componendo and Dividendo rule. This rule states that if \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\) and also \(\frac{a-b}{a+b} = \frac{c-d}{c+d}\) (by inverting the ratio). Alternatively, and more directly useful here, if \(\frac{a-b}{a+b} = \frac{c}{d}\), we can write \(\frac{(a-b)+(a+b)}{(a+b)-(a-b)} = \frac{c+d}{d-c}\).
Let's apply this to our given equation:
\(\frac{{\sec \theta \; - \;\tan \theta }}{{\sec \theta \; + \;\tan \theta }}\; = \;\frac{3}{5}\)
Applying Componendo and Dividendo:
\(\frac{(\sec \theta - \tan \theta) + (\sec \theta + \tan \theta)}{(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta)} = \frac{3+5}{5-3}\)
Simplify the numerator and denominator on the left side:
- Numerator: \((\sec \theta - \tan \theta) + (\sec \theta + \tan \theta) = \sec \theta - \tan \theta + \sec \theta + \tan \theta = 2\sec \theta\)
- Denominator: \((\sec \theta + \tan \theta) - (\sec \theta - \tan \theta) = \sec \theta + \tan \theta - \sec \theta + \tan \theta = 2\tan \theta\)
Simplify the right side:
- Right side: \(\frac{3+5}{5-3} = \frac{8}{2} = 4\)
So, the equation becomes:
\(\frac{2\sec \theta}{2\tan \theta} = 4\)
\(\frac{\sec \theta}{\tan \theta} = 4\)
Finding the Value of Cosecant Theta
Now we have \(\frac{\sec \theta}{\tan \theta} = 4\). We can rewrite \(\sec \theta\) and \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\):
- \(\sec \theta = \frac{1}{\cos \theta}\)
- \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
Substitute these into the equation:
\(\frac{\frac{1}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} = 4\)
\(\frac{1}{\cos \theta} \times \frac{\cos \theta}{\sin \theta} = 4\)
\(\frac{1}{\sin \theta} = 4\)
Since \(cosec \theta = \frac{1}{\sin \theta}\), we have:
\(cosec \theta = 4\)
Finding the Value of Cotangent Theta
We know the value of \(cosec \theta\). We can find \(\cot \theta\) using the fundamental trigonometric identity: \(cosec^2 \theta - \cot^2 \theta = 1\).
Substitute the value of \(cosec \theta = 4\):
\(4^2 - \cot^2 \theta = 1\)
\(16 - \cot^2 \theta = 1\)
Rearrange the equation to solve for \(\cot^2 \theta\):
\(\cot^2 \theta = 16 - 1\)
\(\cot^2 \theta = 15\)
Taking the square root of both sides:
\(\cot \theta = \pm\sqrt{15}\)
Assuming \(\theta\) is in a quadrant where \(\cot \theta\) is positive (e.g., the first quadrant), we take \(\cot \theta = \sqrt{15}\).
Evaluating the Cosecant and Cotangent Expression
Now we need to find the value of \(\frac{{cosec\theta \; + \;\cot \theta }}{{cosec\;\theta \; - {\rm{\;cot}}\theta }}\) using \(cosec \theta = 4\) and \(\cot \theta = \sqrt{15}\).
Substitute the values into the expression:
\(\frac{4 + \sqrt{15}}{4 - \sqrt{15}}\)
To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \(4 + \sqrt{15}\).
\(\frac{4 + \sqrt{15}}{4 - \sqrt{15}} \times \frac{4 + \sqrt{15}}{4 + \sqrt{15}}\)
Multiply the numerators and the denominators:
- Numerator: \((4 + \sqrt{15})(4 + \sqrt{15}) = (4 + \sqrt{15})^2 = 4^2 + 2 \times 4 \times \sqrt{15} + (\sqrt{15})^2 = 16 + 8\sqrt{15} + 15 = 31 + 8\sqrt{15}\)
- Denominator: \((4 - \sqrt{15})(4 + \sqrt{15})\). This is in the form \((a-b)(a+b) = a^2 - b^2\). So, \(4^2 - (\sqrt{15})^2 = 16 - 15 = 1\).
So, the expression becomes:
\(\frac{31 + 8\sqrt{15}}{1} = 31 + 8\sqrt{15}\)
Thus, the value of \(\frac{{cosec\theta \; + \;\cot \theta }}{{cosec\;\theta \; - {\rm{\;cot}}\theta }}\) is \(31 + 8\sqrt{15}\).
Revision Table: Key Trigonometric Identities
| Identity | Description |
|---|---|
| \(\sec \theta = \frac{1}{\cos \theta}\) | Secant is the reciprocal of cosine. |
| \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) | Tangent is the ratio of sine to cosine. |
| \(cosec \theta = \frac{1}{\sin \theta}\) | Cosecant is the reciprocal of sine. |
| \(\cot \theta = \frac{\cos \theta}{\sin \theta}\) | Cotangent is the ratio of cosine to sine (reciprocal of tangent). |
| \(cosec^2 \theta - \cot^2 \theta = 1\) | Pythagorean identity relating cosecant and cotangent. |
| \(\sec^2 \theta - \tan^2 \theta = 1\) | Pythagorean identity relating secant and tangent. |
Additional Information: Componendo and Dividendo Rule
The Componendo and Dividendo rule is a property of proportions. If we have a ratio \(\frac{a}{b} = \frac{c}{d}\), then the rule states that \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\). Similarly, if \(\frac{a-b}{a+b} = \frac{c}{d}\), we can apply the rule as \(\frac{(a-b)+(a+b)}{(a+b)-(a-b)} = \frac{c+d}{d-c}\). This rule is very useful for simplifying equations involving fractions of sums and differences, like the one given in this problem.
In this problem, the form \(\frac{\sec \theta - \tan \theta}{\sec \theta + \tan \theta} = \frac{3}{5}\) fits the pattern \(\frac{a-b}{a+b} = \frac{c}{d}\), where \(a = \sec \theta\), \(b = \tan \theta\), \(c = 3\), and \(d = 5\). Applying the rule helps isolate the ratio \(\frac{a}{b}\) or \(\frac{b}{a}\), which simplifies the problem significantly.
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