What is (1 + cot θ - cosec θ)(1 + tan θ + sec θ) equal to?
2
The question asks us to simplify and find the value of the expression \((1 + \cot \theta - \operatorname{cosec} \theta)(1 + \tan \theta + \sec \theta)\). This involves using fundamental trigonometric identities and algebraic manipulation.
To simplify expressions involving different trigonometric functions like \(\cot \theta\), \(\operatorname{cosec} \theta\), \(\tan \theta\), and \(\sec \theta\), it is often helpful to convert them into their basic forms involving \(\sin \theta\) and \(\cos \theta\).
Let's substitute these equivalent forms into the given expression:
The expression is: \((1 + \cot \theta - \operatorname{cosec} \theta)(1 + \tan \theta + \sec \theta)\)
Substitute the \(\sin \theta\) and \(\cos \theta\) forms:
\(\left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right)\left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right)\)
Now, let's simplify the terms inside each set of parentheses by finding a common denominator:
For the first set of parentheses, the common denominator is \(\sin \theta\):
\(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} = \frac{\sin \theta}{\sin \theta} + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta - 1}{\sin \theta}\)
For the second set of parentheses, the common denominator is \(\cos \theta\):
\(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = \frac{\cos \theta}{\cos \theta} + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = \frac{\cos \theta + \sin \theta + 1}{\cos \theta}\)
Now, multiply these two simplified expressions:
\(\left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right)\left(\frac{\sin \theta + \cos \theta + 1}{\cos \theta}\right)\)
We can rewrite the numerators slightly to group the terms \((\sin \theta + \cos \theta)\):
\(\frac{(\sin \theta + \cos \theta) - 1}{\sin \theta} \times \frac{(\sin \theta + \cos \theta) + 1}{\cos \theta} = \frac{((\sin \theta + \cos \theta) - 1)((\sin \theta + \cos \theta) + 1)}{\sin \theta \cos \theta}\)
The numerator is in the form \((a-b)(a+b)\), where \(a = (\sin \theta + \cos \theta)\) and \(b = 1\). Using the algebraic identity \((a-b)(a+b) = a^2 - b^2\), we get:
Numerator \(= (\sin \theta + \cos \theta)^2 - 1^2\)
Expand \((\sin \theta + \cos \theta)^2\) using \((a+b)^2 = a^2 + b^2 + 2ab\):
\((\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta\)
We know the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\). Substitute this into the expanded term:
\((\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta\)
Now substitute this back into the numerator expression:
Numerator \(= (1 + 2 \sin \theta \cos \theta) - 1\)
Simplify the numerator:
Numerator \(= 1 + 2 \sin \theta \cos \theta - 1 = 2 \sin \theta \cos \theta\)
So, the entire expression becomes:
\(\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}\)
Assuming \(\sin \theta \cos \theta \neq 0\) (which means \(\theta\) is not a multiple of \(\frac{\pi}{2}\)), we can cancel out the term \(\sin \theta \cos \theta\) from the numerator and the denominator:
\(\frac{2 \cancel{\sin \theta \cos \theta}}{\cancel{\sin \theta \cos \theta}} = 2\)
Thus, the value of the expression \((1 + \cot \theta - \operatorname{cosec} \theta)(1 + \tan \theta + \sec \theta)\) is 2.
| Identity Type | Identity |
|---|---|
| Reciprocal Identities |
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| Quotient Identities |
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| Pythagorean Identities |
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When faced with simplifying complex trigonometric expressions, here are some general strategies:
This problem specifically utilized converting to \(\sin \theta\) and \(\cos \theta\), finding common denominators, multiplying fractions, and applying the algebraic identity for difference of squares along with the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\).
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