If p = sec θ - tan θ and q = cosec θ + cot θ, then what is p + q(p - 1) equal to?
-1
Solving the Trigonometric Expression p + q(p - 1)
The problem asks us to find the value of the expression \(p + q(p - 1)\) given the definitions of \(p\) and \(q\) in terms of trigonometric functions.
We are given:
- \(p = \sec \theta - \tan \theta\)
- \(q = \csc \theta + \cot \theta\)
We need to evaluate \(p + q(p - 1)\).
Let's simplify the expressions for \(p\) and \(q\) using the basic identities:
- \(\sec \theta = \frac{1}{\cos \theta}\)
- \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
- \(\csc \theta = \frac{1}{\sin \theta}\)
- \(\cot \theta = \frac{\cos \theta}{\sin \theta}\)
Substituting these into the expressions for \(p\) and \(q\):
For \(p\):
\(p = \sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta} = \frac{1 - \sin \theta}{\cos \theta}\)
For \(q\):
\(q = \csc \theta + \cot \theta = \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta}\)
Now let's evaluate the expression \(p + q(p - 1)\). First, we find \(p - 1\):
\(p - 1 = \frac{1 - \sin \theta}{\cos \theta} - 1 = \frac{1 - \sin \theta - \cos \theta}{\cos \theta}\)
Next, we calculate \(q(p - 1)\):
\(q(p - 1) = \left(\frac{1 + \cos \theta}{\sin \theta}\right) \times \left(\frac{1 - \sin \theta - \cos \theta}{\cos \theta}\right)\)
Multiply the numerators and the denominators:
\(q(p - 1) = \frac{(1 + \cos \theta)(1 - \sin \theta - \cos \theta)}{\sin \theta \cos \theta}\)
Expand the numerator:
\((1 + \cos \theta)(1 - \sin \theta - \cos \theta) = 1(1 - \sin \theta - \cos \theta) + \cos \theta(1 - \sin \theta - \cos \theta)\)
\(= 1 - \sin \theta - \cos \theta + \cos \theta - \sin \theta \cos \theta - \cos^2 \theta\)
\(= 1 - \sin \theta - \sin \theta \cos \theta - \cos^2 \theta\)
So, \(q(p - 1) = \frac{1 - \sin \theta - \sin \theta \cos \theta - \cos^2 \theta}{\sin \theta \cos \theta}\)
Finally, we add \(p\) and \(q(p - 1)\):
\(p + q(p - 1) = \frac{1 - \sin \theta}{\cos \theta} + \frac{1 - \sin \theta - \sin \theta \cos \theta - \cos^2 \theta}{\sin \theta \cos \theta}\)
To add these fractions, we find a common denominator, which is \(\sin \theta \cos \theta\). We multiply the first term's numerator and denominator by \(\sin \theta\):
\(p + q(p - 1) = \frac{(1 - \sin \theta)\sin \theta}{\cos \theta \sin \theta} + \frac{1 - \sin \theta - \sin \theta \cos \theta - \cos^2 \theta}{\sin \theta \cos \theta}\)
\(p + q(p - 1) = \frac{\sin \theta - \sin^2 \theta + 1 - \sin \theta - \sin \theta \cos \theta - \cos^2 \theta}{\sin \theta \cos \theta}\)
Combine like terms in the numerator:
\(p + q(p - 1) = \frac{(\sin \theta - \sin \theta) + (1 - \sin^2 \theta - \cos^2 \theta) - \sin \theta \cos \theta}{\sin \theta \cos \theta}\)
Using the fundamental trigonometric identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have \(1 - \sin^2 \theta - \cos^2 \theta = 1 - (\sin^2 \theta + \cos^2 \theta) = 1 - 1 = 0\).
Substitute this into the numerator:
\(p + q(p - 1) = \frac{0 + 0 - \sin \theta \cos \theta}{\sin \theta \cos \theta}\)
\(p + q(p - 1) = \frac{-\sin \theta \cos \theta}{\sin \theta \cos \theta}\)
Assuming \(\sin \theta \cos \theta \neq 0\), we can cancel the common term:
\(p + q(p - 1) = -1\)
Thus, the value of the expression \(p + q(p - 1)\) is \(-1\).
Revision Table: Key Trigonometric Identities
| Identity Type | Identity |
|---|---|
| Reciprocal Identities | \(\sec \theta = \frac{1}{\cos \theta}\) |
| \(\csc \theta = \frac{1}{\sin \theta}\) | |
| \(\cot \theta = \frac{1}{\tan \theta}\) | |
| Quotient Identities | \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) |
| \(\cot \theta = \frac{\cos \theta}{\sin \theta}\) | |
| Pythagorean Identity | \(\sin^2 \theta + \cos^2 \theta = 1\) |
Additional Information on Evaluating Trigonometric Expressions
Evaluating expressions involving trigonometric functions often requires simplifying them using fundamental trigonometric identities. The key steps usually involve:
- Replacing functions like secant, cosecant, tangent, and cotangent with their sine and cosine equivalents.
- Combining fractions by finding a common denominator.
- Expanding products.
- Using Pythagorean identities (\(\sin^2 \theta + \cos^2 \theta = 1\), \(1 + \tan^2 \theta = \sec^2 \theta\), \(1 + \cot^2 \theta = \csc^2 \theta\)) to simplify squared terms.
- Factoring or cancelling common terms where possible.
Practice with these techniques helps in solving more complex trigonometric problems.
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