If \(\rm \frac{sec\,θ-tan\,θ}{sec\,θ+tan\,θ}=\frac{1}{7}\) , θ lies in first quadrant, then the value of \(\rm \frac{cosec\,\theta+cot^2\,\theta}{cosec\,\theta-cot^2\,\theta}\) is:
Solving Trigonometric Expression with Secant, Tangent, Cosecant, and Cotangent
We are given a trigonometric equation involving secant and tangent, and we need to find the value of another expression involving cosecant and cotangent. The angle \(\theta\) is specified to be in the first quadrant.
The given equation is:
\(\rm \frac{sec\,θ-tan\,θ}{sec\,θ+tan\,θ}=\frac{1}{7}\)
We can simplify this equation to find the values of the trigonometric ratios. One effective way is to use the property of Componendo and Dividendo.
Applying Componendo and Dividendo
The Componendo and Dividendo rule states that if \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\). Applying this to our equation:
\(\frac{(sec\,θ-tan\,θ) + (sec\,θ+tan\,θ)}{(sec\,θ+tan\,θ) - (sec\,θ-tan\,θ)} = \frac{1+7}{7-1}\)
Simplifying the numerator and the denominator on the left side:
Numerator: \(sec\,\theta - tan\,\theta + sec\,\theta + tan\,\theta = 2\,sec\,\theta\)
Denominator: \(sec\,\theta + tan\,\theta - sec\,\theta + tan\,\theta = 2\,tan\,\theta\)
Simplifying the right side:
\(\frac{1+7}{7-1} = \frac{8}{6} = \frac{4}{3}\)
So, the equation becomes:
\(\frac{2\,sec\,\theta}{2\,tan\,\theta} = \frac{4}{3}\)
\(\frac{sec\,\theta}{tan\,\theta} = \frac{4}{3}\)
Finding Sine and Cosine Values
We know that \(sec\,\theta = \frac{1}{cos\,\theta}\) and \(tan\,\theta = \frac{sin\,\theta}{cos\,\theta}\). Substituting these into the simplified equation:
\(\frac{\frac{1}{cos\,\theta}}{\frac{sin\,\theta}{cos\,\theta}} = \frac{4}{3}\)
\(\frac{1}{cos\,\theta} \times \frac{cos\,\theta}{sin\,\theta} = \frac{4}{3}\)
\(\frac{1}{sin\,\theta} = \frac{4}{3}\)
This directly gives us the value of \(sin\,\theta\):
\(sin\,\theta = \frac{3}{4}\)
Since \(\theta\) lies in the first quadrant, both \(sin\,\theta\) and \(cos\,\theta\) are positive. We can find \(cos\,\theta\) using the identity \(sin^2\,\theta + cos^2\,\theta = 1\):
\(cos^2\,\theta = 1 - sin^2\,\theta\)
\(cos^2\,\theta = 1 - \left(\frac{3}{4}\right)^2\)
\(cos^2\,\theta = 1 - \frac{9}{16}\)
\(cos^2\,\theta = \frac{16-9}{16} = \frac{7}{16}\)
\(cos\,\theta = \sqrt{\frac{7}{16}}\) (Since \(\theta\) is in the first quadrant, \(cos\,\theta > 0\))
\(cos\,\theta = \frac{\sqrt{7}}{4}\)
Calculating Cosecant and Cotangent
Now we need to find the values of \(cosec\,\theta\) and \(cot\,\theta\) to evaluate the required expression. We know that:
\(cosec\,\theta = \frac{1}{sin\,\theta}\)
\(cosec\,\theta = \frac{1}{3/4} = \frac{4}{3}\)
And \(cot\,\theta = \frac{cos\,\theta}{sin\,\theta}\):
\(cot\,\theta = \frac{\sqrt{7}/4}{3/4} = \frac{\sqrt{7}}{4} \times \frac{4}{3} = \frac{\sqrt{7}}{3}\)
We also need \(cot^2\,\theta\):
\(cot^2\,\theta = \left(\frac{\sqrt{7}}{3}\right)^2 = \frac{(\sqrt{7})^2}{3^2} = \frac{7}{9}\)
Evaluating the Final Expression
The expression we need to evaluate is \(\rm \frac{cosec\,\theta+cot^2\,\theta}{cosec\,\theta-cot^2\,\theta}\). Substitute the values we found for \(cosec\,\theta\) and \(cot^2\,\theta\):
\(\frac{\frac{4}{3} + \frac{7}{9}}{\frac{4}{3} - \frac{7}{9}}\)
To simplify this complex fraction, find a common denominator for the terms in the numerator and the denominator, which is 9.
Numerator: \(\frac{4}{3} + \frac{7}{9} = \frac{4 \times 3}{3 \times 3} + \frac{7}{9} = \frac{12}{9} + \frac{7}{9} = \frac{12+7}{9} = \frac{19}{9}\)
Denominator: \(\frac{4}{3} - \frac{7}{9} = \frac{4 \times 3}{3 \times 3} - \frac{7}{9} = \frac{12}{9} - \frac{7}{9} = \frac{12-7}{9} = \frac{5}{9}\)
Now substitute these back into the main expression:
\(\frac{\frac{19}{9}}{\frac{5}{9}}\)
To divide fractions, multiply the numerator by the reciprocal of the denominator:
\(\frac{19}{9} \times \frac{9}{5} = \frac{19}{5}\)
Thus, the value of the expression \(\rm \frac{cosec\,\theta+cot^2\,\theta}{cosec\,\theta-cot^2\,\theta}\) is \(\frac{19}{5}\).
Revision Table: Key Trigonometric Values
| Trigonometric Ratio | Value |
|---|---|
| \(sin\,\theta\) | \(\frac{3}{4}\) |
| \(cos\,\theta\) | \(\frac{\sqrt{7}}{4}\) |
| \(cosec\,\theta\) | \(\frac{4}{3}\) |
| \(cot\,\theta\) | \(\frac{\sqrt{7}}{3}\) |
| \(cot^2\,\theta\) | \(\frac{7}{9}\) |
Additional Information: Trigonometric Identities and Properties
This problem utilized several key trigonometric concepts:
- Reciprocal Identities: \(sec\,\theta = \frac{1}{cos\,\theta}\), \(cosec\,\theta = \frac{1}{sin\,\theta}\), \(cot\,\theta = \frac{1}{tan\,\theta}\).
- Quotient Identity: \(tan\,\theta = \frac{sin\,\theta}{cos\,\theta}\), \(cot\,\theta = \frac{cos\,\theta}{sin\,\theta}\).
- Pythagorean Identity: \(sin^2\,\theta + cos^2\,\theta = 1\). Other related identities are \(sec^2\,\theta - tan^2\,\theta = 1\) and \(cosec^2\,\theta - cot^2\,\theta = 1\).
- Quadrant Information: Knowing the quadrant (\(\theta\) is in the first quadrant) helps determine the sign of the trigonometric ratios. In the first quadrant, all trigonometric ratios are positive.
- Componendo and Dividendo: This is a useful algebraic technique that can simplify ratios efficiently. If \(\frac{a}{b} = \frac{c}{d}\), then \(\frac{a+b}{a-b} = \frac{c+d}{c-d}\) and \(\frac{a-b}{a+b} = \frac{c-d}{c+d}\).
Understanding these identities and techniques is crucial for solving various trigonometry problems.
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