If \(cosec~\theta =\frac{29}{21}\) where 0 < θ < 90°, then what is the value of 4 sec θ + 4 tan θ?
10
Understanding the Trigonometry Problem
The question asks us to find the value of an expression involving trigonometric ratios, specifically 4 sec θ + 4 tan θ, given the value of cosec θ and the range for the angle θ.
We are given:
- \(cosec~\theta = \frac{29}{21}\)
- 0 < θ < 90°
The condition 0 < θ < 90° tells us that θ lies in the first quadrant. In the first quadrant, all trigonometric ratios (sine, cosine, tangent, cosecant, secant, cotangent) are positive. This is an important piece of information when dealing with square roots later.
Key Trigonometric Ratios and Identities
To solve this problem, we need to relate the given cosec θ to sec θ and tan θ. We can do this using the definitions of these ratios in terms of the sides of a right-angled triangle, or by using trigonometric identities.
Let's use the right-angled triangle approach:
- \(cosec~\theta = \frac{Hypotenuse}{Opposite~Side}\)
- \(sec~\theta = \frac{Hypotenuse}{Adjacent~Side}\)
- \(tan~\theta = \frac{Opposite~Side}{Adjacent~Side}\)
We know \(cosec~\theta = \frac{29}{21}\). We can consider a right-angled triangle where the Hypotenuse is 29 units and the Opposite Side to angle θ is 21 units. Let the Adjacent Side be 'a'.
Using the Pythagorean theorem (\(Hypotenuse^2 = Opposite^2 + Adjacent^2\)):
\(29^2 = 21^2 + a^2\) \(841 = 441 + a^2\) \(a^2 = 841 - 441\) \(a^2 = 400\) \(a = \sqrt{400}\)Since θ is in the first quadrant, the adjacent side is positive.
\(a = 20\)So, the sides of the right-angled triangle can be considered as:
- Hypotenuse = 29
- Opposite Side = 21
- Adjacent Side = 20
Step-by-Step Solution
Now that we have the lengths of all sides of the right-angled triangle (in proportion), we can find the values of sec θ and tan θ.
Step 1: Find sec θ
Using the definition \(sec~\theta = \frac{Hypotenuse}{Adjacent~Side}\):
\(sec~\theta = \frac{29}{20}\)Step 2: Find tan θ
Using the definition \(tan~\theta = \frac{Opposite~Side}{Adjacent~Side}\):
\(tan~\theta = \frac{21}{20}\)Step 3: Calculate the value of 4 sec θ + 4 tan θ
Substitute the values of sec θ and tan θ into the expression:
\(4~sec~\theta + 4~tan~\theta = 4 \left(\frac{29}{20}\right) + 4 \left(\frac{21}{20}\right)\) \(= \frac{4 \times 29}{20} + \frac{4 \times 21}{20}\) \(= \frac{116}{20} + \frac{84}{20}\)Since the denominators are the same, we can add the numerators:
\(= \frac{116 + 84}{20}\) \(= \frac{200}{20}\) \(= 10\)Calculating the Final Value
The value of 4 sec θ + 4 tan θ is 10.
| Trigonometric Ratio | Value from Triangle |
|---|---|
| \(cosec~\theta\) | \(\frac{29}{21}\) (Given) |
| \(sec~\theta\) | \(\frac{29}{20}\) |
| \(tan~\theta\) | \(\frac{21}{20}\) |
Substituting these values into the expression \(4 sec~\theta + 4 tan~\theta\):
\(4 \left(\frac{29}{20}\right) + 4 \left(\frac{21}{20}\right) = \frac{116}{20} + \frac{84}{20} = \frac{200}{20} = 10\)Revision Table: Key Trigonometric Ratios
| Ratio | Definition (Right Triangle) | Reciprocal |
|---|---|---|
| Sine (\(\sin \theta\)) | \(\frac{Opposite}{Hypotenuse}\) | \(cosec~\theta = \frac{1}{\sin \theta}\) |
| Cosine (\(\cos \theta\)) | \(\frac{Adjacent}{Hypotenuse}\) | \(sec~\theta = \frac{1}{\cos \theta}\) |
| Tangent (\(\tan \theta\)) | \(\frac{Opposite}{Adjacent}\) | \(cot~\theta = \frac{1}{\tan \theta}\) |
Additional Information: Trigonometric Identities
Alternatively, we could use trigonometric identities. We are given \(cosec~\theta = \frac{29}{21}\).
We know that \(sin~\theta = \frac{1}{cosec~\theta}\).
\(sin~\theta = \frac{1}{\frac{29}{21}} = \frac{21}{29}\)Using the identity \(sin^2 \theta + cos^2 \theta = 1\):
\(\left(\frac{21}{29}\right)^2 + cos^2 \theta = 1\) \(\frac{441}{841} + cos^2 \theta = 1\) \(cos^2 \theta = 1 - \frac{441}{841} = \frac{841 - 441}{841} = \frac{400}{841}\)Since θ is in the first quadrant, \(cos~\theta\) is positive.
\(cos~\theta = \sqrt{\frac{400}{841}} = \frac{20}{29}\)Now, find \(sec~\theta\) and \(tan~\theta\):
\(sec~\theta = \frac{1}{cos~\theta} = \frac{1}{\frac{20}{29}} = \frac{29}{20}\) \(tan~\theta = \frac{sin~\theta}{cos~\theta} = \frac{\frac{21}{29}}{\frac{20}{29}} = \frac{21}{20}\)Substitute these values into the expression \(4 sec~\theta + 4 tan~\theta\):
\(4 \left(\frac{29}{20}\right) + 4 \left(\frac{21}{20}\right) = \frac{116}{20} + \frac{84}{20} = \frac{200}{20} = 10\)Both methods yield the same result, confirming the answer.
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