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.
Similar Questions
What is the value of sin 26° + sin 212° + sin 218° + … + sin 284° + sin 290°?
Consider the following statements:
1. (sec 2θ - 1) (1 - cosec 2θ) = 1
2. sin θ (1 + cos θ) -1 + (1 + cos θ) (sin θ) -1 = 2 cosec θ
Which of the above is/are correct?
Important Questions from Trigonometric Identities
If cos 2θ = sin θ and θ lies between 0 and 90°, then θ will be:
\(\frac{3 - 4\sin^2 \theta}{\cos^2 \theta} + 2\tan^2 \theta\) can be simplified as:
Simplify \( \left(\frac{1}{\sin^2 A} - 1\right) \), where \( 0 < A \leq 90^\circ \).
If \( \tan \theta = \frac{8}{15} \), then the value of \( \sqrt{\frac{1 - \sin \theta}{1 + \sin \theta}} \) is:
If the volume of a cuboid is \(3x^2 - 27\), then its possible dimensions are:
