Simplify \(\frac{1+sint}{4-4sint}-\frac{1-sint}{4+4sint}\)
tan t . sec t
The problem asks us to simplify the given trigonometric expression:
\[ \frac{1+\sin t}{4-4\sin t} - \frac{1-\sin t}{4+4\sin t} \]
To simplify this expression, we will first find a common denominator for the two fractions.
The denominators are \(4-4\sin t\) and \(4+4\sin t\). We can factor out 4 from each denominator:
The common denominator is \(4(1-\sin t) \times (1+\sin t)\). Using the difference of squares formula, \((a-b)(a+b) = a^2 - b^2\), we get:
\[ 4(1-\sin t)(1+\sin t) = 4(1^2 - \sin^2 t) = 4(1 - \sin^2 t) \]
Recall the fundamental trigonometric identity \( \sin^2 t + \cos^2 t = 1 \), which implies \( 1 - \sin^2 t = \cos^2 t \). So, the common denominator is \( 4\cos^2 t \).
Alternatively, we can simply use \((4-4\sin t)(4+4\sin t)\) as the common denominator initially and simplify later:
\[ (4-4\sin t)(4+4\sin t) = 4^2 - (4\sin t)^2 = 16 - 16\sin^2 t = 16(1 - \sin^2 t) = 16\cos^2 t \]
Let's use the common denominator \(16\cos^2 t\).
Now, we rewrite each fraction with the common denominator:
\[ \frac{1+\sin t}{4-4\sin t} = \frac{(1+\sin t)(4+4\sin t)}{(4-4\sin t)(4+4\sin t)} = \frac{4+4\sin t+4\sin t+4\sin^2 t}{16\cos^2 t} = \frac{4+8\sin t+4\sin^2 t}{16\cos^2 t} \]
\[ \frac{1-\sin t}{4+4\sin t} = \frac{(1-\sin t)(4-4\sin t)}{(4+4\sin t)(4-4\sin t)} = \frac{4-4\sin t-4\sin t+4\sin^2 t}{16\cos^2 t} = \frac{4-8\sin t+4\sin^2 t}{16\cos^2 t} \]
Now, subtract the second expression from the first:
\[ \left(\frac{4+8\sin t+4\sin^2 t}{16\cos^2 t}\right) - \left(\frac{4-8\sin t+4\sin^2 t}{16\cos^2 t}\right) \]
Combine the numerators over the common denominator:
\[ \frac{(4+8\sin t+4\sin^2 t) - (4-8\sin t+4\sin^2 t)}{16\cos^2 t} \]
Simplify the numerator:
\[ 4+8\sin t+4\sin^2 t - 4+8\sin t-4\sin^2 t = (4-4) + (8\sin t+8\sin t) + (4\sin^2 t-4\sin^2 t) = 16\sin t \]
So the expression simplifies to:
\[ \frac{16\sin t}{16\cos^2 t} \]
Cancel out the common factor of 16 from the numerator and denominator:
\[ \frac{\sin t}{\cos^2 t} \]
We can rewrite this expression using the definitions of tangent and secant functions:
\[ \frac{\sin t}{\cos^2 t} = \frac{\sin t}{\cos t \cdot \cos t} = \frac{\sin t}{\cos t} \cdot \frac{1}{\cos t} \]
Recall that \( \tan t = \frac{\sin t}{\cos t} \) and \( \sec t = \frac{1}{\cos t} \).
Therefore, the simplified expression is:
\[ \tan t \cdot \sec t \]
This matches one of the given options.
| Original Expression | Step | Result |
|---|---|---|
| \( \frac{1+\sin t}{4-4\sin t} - \frac{1-\sin t}{4+4\sin t} \) | Find common denominator \(16(1-\sin^2 t) = 16\cos^2 t\) | \( \frac{(1+\sin t)(4+4\sin t) - (1-\sin t)(4-4\sin t)}{16\cos^2 t} \) |
| Expand and simplify numerator | \( \frac{(4+8\sin t+4\sin^2 t) - (4-8\sin t+4\sin^2 t)}{16\cos^2 t} = \frac{16\sin t}{16\cos^2 t} \) | |
| Cancel common factor 16 | \( \frac{\sin t}{\cos^2 t} \) | |
| Rewrite using \( \tan t \) and \( \sec t \) definitions | \( \frac{\sin t}{\cos t} \cdot \frac{1}{\cos t} = \tan t \cdot \sec t \) |
| Concept | Formula/Identity | Notes |
|---|---|---|
| Difference of Squares | \( a^2 - b^2 = (a-b)(a+b) \) | Used for common denominator |
| Pythagorean Identity | \( \sin^2 \theta + \cos^2 \theta = 1 \) | Leads to \( 1 - \sin^2 \theta = \cos^2 \theta \) |
| Tangent Definition | \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) | Ratio of sine to cosine |
| Secant Definition | \( \sec \theta = \frac{1}{\cos \theta} \) | Reciprocal of cosine |
Simplifying trigonometric expressions often involves using fundamental identities and algebraic techniques like finding common denominators, factoring, and expanding terms. The goal is usually to express the result in terms of fewer trigonometric functions or a standard form.
Key strategies include:
Practicing with various expressions helps in recognizing which identity or technique is most useful for a particular problem.
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