Simplify Trigonometric Expression
We are given the equation: \( \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = p \sec \theta + q \tan \theta \) Our goal is to simplify the left-hand side (LHS) of the equation.
First, let's rearrange the terms in the numerator and the denominator:
\( \text{LHS} = \frac{(1 + \sin \theta) - \cos \theta}{(\sin \theta - 1) + \cos \theta} \)To simplify this expression, we can divide both the numerator and the denominator by \(\cos \theta\) (since \(0 < \theta < \frac{\pi}{2}\), \(\cos \theta \neq 0\)):
\( \text{LHS} = \frac{\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta} - \frac{1}{\cos \theta}} \)Using the identities \(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we get:
\( \text{LHS} = \frac{\sec \theta + \tan \theta - 1}{\tan \theta + 1 - \sec \theta} \)Now, let's rearrange the terms again:
\( \text{LHS} = \frac{(\sec \theta + \tan \theta) - 1}{1 - \sec \theta + \tan \theta} \)We can use the trigonometric identity \(1 = \sec^2 \theta - \tan^2 \theta\). Substitute this into the numerator:
\( \text{LHS} = \frac{\sec \theta + \tan \theta - (\sec^2 \theta - \tan^2 \theta)}{\tan \theta + 1 - \sec \theta} \)Factor the difference of squares term \(\sec^2 \theta - \tan^2 \theta = (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)\):
\( \text{LHS} = \frac{(\sec \theta + \tan \theta) - (\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\tan \theta + 1 - \sec \theta} \)Factor out \((\sec \theta + \tan \theta)\) from the numerator:
\( \text{LHS} = \frac{(\sec \theta + \tan \theta) [1 - (\sec \theta - \tan \theta)]}{\tan \theta + 1 - \sec \theta} \) \( \text{LHS} = \frac{(\sec \theta + \tan \theta) (1 - \sec \theta + \tan \theta)}{\tan \theta + 1 - \sec \theta} \)Notice that the term \((1 - \sec \theta + \tan \theta)\) is the same as \((\tan \theta + 1 - \sec \theta)\). Since \(0 < \theta < \frac{\pi}{2}\), the denominator \(\tan \theta + 1 - \sec \theta\) is not zero. Thus, we can cancel this term from the numerator and denominator:
\( \text{LHS} = \sec \theta + \tan \theta \)Equate and Compare Coefficients
Now, we equate the simplified LHS with the given right-hand side (RHS):
\( \sec \theta + \tan \theta = p \sec \theta + q \tan \theta \)For this equation to hold true for all \(\theta\) in the specified range, the coefficients of \(\sec \theta\) and \(\tan \theta\) on both sides must be equal.
- Comparing the coefficients of \(\sec \theta\): \(1 = p\)
- Comparing the coefficients of \(\tan \theta\): \(1 = q\)
Calculate the Final Value
We found that \(p = 1\) and \(q = 1\). The question asks for the value of \(p + q\).
\( p + q = 1 + 1 = 2 \)Therefore, the value of \(p + q\) is 2.
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Important Questions from Trigonometric Identities
What is \(\rm \frac{1+tan^2\theta}{1+cot^2\theta}-\left(\frac{1-tan\theta}{1-cot\theta}\right)^2\) equal to?
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If \(\sin \left( {A - B} \right) = \frac{1}{2}\) and \(\cos \left( {A + B} \right) = \frac{1}{2}\) , where A > B > 0° and A + B is an acute angle, then the value of A is:
Find the value of $\cos 10^\circ \times \cos 30^\circ \times \cos 50^\circ \times \cos 70^\circ$
