For the following two (02) items : Let $\text{cosec}\theta - \sin\theta = p$ and $\sec\theta - \cos\theta = q$.
Understanding the Trigonometric Expressions
We are given two initial expressions involving the angle \(\theta\):
- \(p = \text{cosec}\theta - \sin\theta\)
- \(q = \sec\theta - \cos\theta\)
The task is to determine the value of the expression \((p\sin\theta + q\cos\theta)\).
Simplifying the Expression for p
First, let's simplify the expression for \(p\). We know that \(\text{cosec}\theta\) is the reciprocal of \(\sin\theta\), meaning \(\text{cosec}\theta = \frac{1}{\sin\theta}\). Substituting this into the expression for \(p\):
\(p = \frac{1}{\sin\theta} - \sin\theta\)
To combine these terms, we use a common denominator, \(\sin\theta\):
\(p = \frac{1 - \sin^2\theta}{\sin\theta}\)
Recall the fundamental Pythagorean trigonometric identity: \(\sin^2\theta + \cos^2\theta = 1\). Rearranging this gives \(1 - \sin^2\theta = \cos^2\theta\). Substituting this identity:
\(p = \frac{\cos^2\theta}{\sin\theta}\)
Simplifying the Expression for q
Next, we simplify the expression for \(q\). We know that \(\sec\theta\) is the reciprocal of \(\cos\theta\), meaning \(\sec\theta = \frac{1}{\cos\theta}\). Substituting this into the expression for \(q\):
\(q = \frac{1}{\cos\theta} - \cos\theta\)
Using a common denominator, \(\cos\theta\):
\(q = \frac{1 - \cos^2\theta}{\cos\theta}\)
Again, using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\), we can rearrange it to \(1 - \cos^2\theta = \sin^2\theta\). Substituting this identity:
\(q = \frac{\sin^2\theta}{\cos\theta}\)
Evaluating the Target Expression \((p\sin\theta + q\cos\theta)\)
Now we substitute the simplified forms of \(p\) and \(q\) into the expression we need to evaluate:
\((p\sin\theta + q\cos\theta) = \left(\frac{\cos^2\theta}{\sin\theta}\right) \sin\theta + \left(\frac{\sin^2\theta}{\cos\theta}\right) \cos\theta\)
We can see that \(\sin\theta\) cancels in the first term and \(\cos\theta\) cancels in the second term:
\((p\sin\theta + q\cos\theta) = \cos^2\theta + \sin^2\theta\)
Finally, we apply the fundamental Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\) one last time:
\((p\sin\theta + q\cos\theta) = 1\)
Thus, the value of the expression \((p\sin\theta + q\cos\theta)\) is 1.
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