The expression \(\rm \frac{(1-sin\,θ+cos\,θ)^2(1-cos\,θ)sec^3\,θ cosec^2\,θ}{(sec\,θ -tan\,θ)(tan\,θ+cot\,θ)} \) , 0° < θ < 90°, is equal to:
2 tan θ
Simplifying Trigonometric Expressions
Let's simplify the given trigonometric expression step-by-step. The expression is:
\(\rm \frac{(1-sin\,θ+cos\,θ)^2(1-cos\,θ)sec^3\,θ cosec^2\,θ}{(sec\,θ -tan\,θ)(tan\,θ+cot\,θ)} \)
where \(0° < θ < 90°\).
Step 1: Simplify the Numerator
The numerator is \(\rm (1-sin\,θ+cos\,θ)^2(1-cos\,θ)sec^3\,θ cosec^2\,θ \).
Consider the term \(\rm (1-sin\,θ+cos\,θ)^2\). We can group terms: \(\rm ((1+cos\,θ) - sin\,θ)^2 \). Expanding this gives:
\(\rm (1+cos\,θ)^2 - 2(1+cos\,θ)sin\,θ + sin^2\,θ \)
\(\rm = (1 + 2cos\,θ + cos^2\,θ) - (2sin\,θ + 2sin\,θcos\,θ) + sin^2\,θ \)
Using the identity \(\rm sin^2\,θ + cos^2\,θ = 1\):
\(\rm = 1 + 2cos\,θ + 1 - 2sin\,θ - 2sin\,θcos\,θ \)
\(\rm = 2 + 2cos\,θ - 2sin\,θ - 2sin\,θcos\,θ \)
\(\rm = 2(1 + cos\,θ - sin\,θ - sin\,θcos\,θ) \)
Factor by grouping:
\(\rm = 2(1(1 + cos\,θ) - sin\,θ(1 + cos\,θ)) \)
\(\rm = 2(1 - sin\,θ)(1 + cos\,θ) \)
Now substitute this back into the numerator:
\(\rm \text{Numerator} = 2(1 - sin\,θ)(1 + cos\,θ)(1-cos\,θ)sec^3\,θ cosec^2\,θ \)
Using the identity \(\rm (1 + cos\,θ)(1-cos\,θ) = 1 - cos^2\,θ = sin^2\,θ \):
\(\rm \text{Numerator} = 2(1 - sin\,θ)sin^2\,θ sec^3\,θ cosec^2\,θ \)
Replace \(\rm sec\,θ = \frac{1}{cos\,θ} \) and \(\rm cosec\,θ = \frac{1}{sin\,θ} \):
\(\rm \text{Numerator} = 2(1 - sin\,θ)sin^2\,θ \left(\frac{1}{cos^3\,θ}\right) \left(\frac{1}{sin^2\,θ}\right) \)
Cancel the \(\rm sin^2\,θ \) terms (since \(0° < θ < 90°\), \(\rm sin\,θ \neq 0 \)):
\(\rm \text{Numerator} = \frac{2(1 - sin\,θ)}{cos^3\,θ} \)
Step 2: Simplify the Denominator
The denominator is \(\rm (sec\,θ -tan\,θ)(tan\,θ+cot\,θ) \).
Replace terms using \(\rm sec\,θ = \frac{1}{cos\,θ} \), \(\rm tan\,θ = \frac{sin\,θ}{cos\,θ} \), and \(\rm cot\,θ = \frac{cos\,θ}{sin\,θ} \):
\(\rm sec\,θ -tan\,θ = \frac{1}{cos\,θ} - \frac{sin\,θ}{cos\,θ} = \frac{1 - sin\,θ}{cos\,θ} \)
\(\rm tan\,θ+cot\,θ = \frac{sin\,θ}{cos\,θ} + \frac{cos\,θ}{sin\,θ} = \frac{sin^2\,θ + cos^2\,θ}{sin\,θ cos\,θ} \)
Using the identity \(\rm sin^2\,θ + cos^2\,θ = 1\):
\(\rm tan\,θ+cot\,θ = \frac{1}{sin\,θ cos\,θ} \)
Now multiply these two simplified parts for the denominator:
\(\rm \text{Denominator} = \left(\frac{1 - sin\,θ}{cos\,θ}\right) \left(\frac{1}{sin\,θ cos\,θ}\right) \)
\(\rm \text{Denominator} = \frac{1 - sin\,θ}{sin\,θ cos^2\,θ} \)
Step 3: Divide the Simplified Numerator by the Simplified Denominator
The expression is \(\rm \frac{\text{Numerator}}{\text{Denominator}} \).
\(\rm \frac{\frac{2(1 - sin\,θ)}{cos^3\,θ}}{\frac{1 - sin\,θ}{sin\,θ cos^2\,θ}} \)
To divide, multiply the numerator by the reciprocal of the denominator:
\(\rm = \frac{2(1 - sin\,θ)}{cos^3\,θ} \times \frac{sin\,θ cos^2\,θ}{1 - sin\,θ} \)
Since \(0° < θ < 90°\), \(\rm sin\,θ \) is between 0 and 1, so \(\rm 1 - sin\,θ \neq 0 \). We can cancel the \(\rm (1 - sin\,θ) \) terms:
\(\rm = \frac{2}{cos^3\,θ} \times \frac{sin\,θ cos^2\,θ}{1} \)
\(\rm = \frac{2 sin\,θ cos^2\,θ}{cos^3\,θ} \)
Cancel the common factors of \(\rm cos^2\,θ \) (since \(0° < θ < 90°\), \(\rm cos\,θ \neq 0 \)):
\(\rm = \frac{2 sin\,θ}{cos\,θ} \)
Using the identity \(\rm \frac{sin\,θ}{cos\,θ} = tan\,θ \):
\(\rm = 2 tan\,θ \)
Conclusion
The simplified expression is \(\rm 2 tan\,θ \).
| Original Expression Part | Simplified Form | Identities Used |
|---|---|---|
| \(\rm (1-sin\,θ+cos\,θ)^2\) | \(\rm 2(1 - sin\,θ)(1 + cos\,θ)\) | Algebraic expansion, \(\rm sin^2 θ + cos^2 θ = 1\) |
| \(\rm (1 + cos\,θ)(1-cos\,θ)\) | \(\rm sin^2 θ \) | \(\rm 1 - cos^2 θ = sin^2 θ \) |
| \(\rm sec^3\,θ \) | \(\rm 1/cos^3 θ \) | \(\rm sec\,θ = 1/cos\,θ \) |
| \(\rm cosec^2\,θ \) | \(\rm 1/sin^2 θ \) | \(\rm cosec\,θ = 1/sin\,θ \) |
| \(\rm sec\,θ -tan\,θ \) | \(\rm (1 - sin\,θ)/cos\,θ \) | \(\rm sec\,θ = 1/cos\,θ, tan\,θ = sin\,θ/cos\,θ \) |
| \(\rm tan\,θ+cot\,θ \) | \(\rm 1/(sin\,θ cos\,θ) \) | \(\rm tan\,θ = sin\,θ/cos\,θ, cot\,θ = cos\,θ/sin\,θ, sin^2 θ + cos^2 θ = 1\) |
Revision Table: Key Trigonometric Identities
| Identity | Description |
|---|---|
| \(\rm sin^2 θ + cos^2 θ = 1 \) | Pythagorean identity |
| \(\rm sec\,θ = \frac{1}{cos\,θ} \) | Reciprocal identity |
| \(\rm cosec\,θ = \frac{1}{sin\,θ} \) | Reciprocal identity |
| \(\rm tan\,θ = \frac{sin\,θ}{cos\,θ} \) | Quotient identity |
| \(\rm cot\,θ = \frac{cos\,θ}{sin\,θ} \) | Quotient identity |
| \(\rm cot\,θ = \frac{1}{tan\,θ} \) | Reciprocal identity |
Additional Information on Trigonometric Simplification
Simplifying trigonometric expressions often involves using fundamental identities to rewrite parts of the expression in terms of sine and cosine. Then, algebraic techniques like factoring, expanding, and finding common denominators are applied. It is crucial to remember the domain of the variables, as it might affect which terms can be cancelled (e.g., division by zero). In this problem, the condition \(0° < θ < 90°\) ensures that \(\rm sin\,θ \neq 0 \) and \(\rm cos\,θ \neq 0 \), and also \(1 - sin\,θ \neq 0 \), allowing cancellations.
Key strategies include:
- Converting all terms to sine and cosine.
- Using Pythagorean identities (\(\rm sin^2 θ + cos^2 θ = 1 \), \(\rm 1 + tan^2 θ = sec^2 θ \), \(\rm 1 + cot^2 θ = cosec^2 θ \)).
- Factoring expressions.
- Finding a common denominator for fractions.
- Recognizing algebraic patterns like \(\rm (a-b)^2 \) or \(\rm (a-b)(a+b) \).
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