For the following two (02) items: Let $\frac{1+\sin\theta}{\cos\theta} = p + \sqrt{p^2+1}$.
Trigonometry: Simplifying the Expression for sec(\(\theta\))
The problem asks us to find the value of \(\sec\theta\) given a specific trigonometric equation involving \(\sin\theta\), \(\cos\theta\), and a variable \(p\). The given equation is:
\(\frac{1+\sin\theta}{\cos\theta} = p + \sqrt{p^2+1}\)
Step-by-Step Solution Breakdown
We can simplify the left side of the given equation:
\(\frac{1+\sin\theta}{\cos\theta} = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\)
Using the basic trigonometric identities \(\sec\theta = \frac{1}{\cos\theta}\) and \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), we get:
\(\sec\theta + \tan\theta = p + \sqrt{p^2+1}\)
Let's call this Equation (1).
Now, we use another fundamental trigonometric identity: \(\sec^2\theta - \tan^2\theta = 1\). This identity can be factored as a difference of squares:
\((\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1\)
We already know the value of \((\sec\theta + \tan\theta)\) from Equation (1). Substitute it into the factored identity:
\((\sec\theta - \tan\theta)(p + \sqrt{p^2+1}) = 1\)
Now, we can solve for \((\sec\theta - \tan\theta)\):
\(\sec\theta - \tan\theta = \frac{1}{p + \sqrt{p^2+1}}\)
To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \((p - \sqrt{p^2+1})\):
\(\sec\theta - \tan\theta = \frac{1}{p + \sqrt{p^2+1}} \times \frac{p - \sqrt{p^2+1}}{p - \sqrt{p^2+1}}\)
\(\sec\theta - \tan\theta = \frac{p - \sqrt{p^2+1}}{p^2 - (\sqrt{p^2+1})^2}\)
\(\sec\theta - \tan\theta = \frac{p - \sqrt{p^2+1}}{p^2 - (p^2+1)}\)
\(\sec\theta - \tan\theta = \frac{p - \sqrt{p^2+1}}{p^2 - p^2 - 1}\)
\(\sec\theta - \tan\theta = \frac{p - \sqrt{p^2+1}}{-1}\)
\(\sec\theta - \tan\theta = -(p - \sqrt{p^2+1})\)
\(\sec\theta - \tan\theta = \sqrt{p^2+1} - p\)
Let's call this Equation (2).
Finding the Value of sec(\(\theta\))
Now we have a system of two linear equations with \(\sec\theta\) and \(\tan\theta\):
| Equation | Expression |
| (1) | \(\sec\theta + \tan\theta = p + \sqrt{p^2+1}\) |
| (2) | \(\sec\theta - \tan\theta = \sqrt{p^2+1} - p\) |
To find \(\sec\theta\), we can add Equation (1) and Equation (2) together:
\((\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = (p + \sqrt{p^2+1}) + (\sqrt{p^2+1} - p)\)
The \(\tan\theta\) terms cancel out:
\(\sec\theta + \sec\theta + \tan\theta - \tan\theta = p + \sqrt{p^2+1} + \sqrt{p^2+1} - p\)
\(2\sec\theta = 2\sqrt{p^2+1}\)
Finally, divide both sides by 2 to isolate \(\sec\theta\):
\(\sec\theta = \frac{2\sqrt{p^2+1}}{2}\)
\(\sec\theta = \sqrt{p^2+1}\)
Conclusion
By simplifying the given trigonometric expression and utilizing the Pythagorean identity \(\sec^2\theta - \tan^2\theta = 1\), we determined that \(\sec\theta\) is equal to \(\sqrt{p^2+1}\).
Similar Questions
If \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} = \text{cosec}\theta\), then what is \(\sin\theta\) equal to?
- If \(2 \tan\theta = \sec^2\theta-2\), where \(0 < \theta < \pi /2\), then what is \(\cot\theta\) equal to?
- What is \(\frac{\tan^2 \alpha}{\tan^2 \beta}\) equal to?
- If \(\alpha\) and \(\beta\) are complementary angles, then which one of the following is correct?
- What is \(\tan\theta\) equal to?
- What is the relation between \(p\) and \(q\)?
- What is $\tan\theta + \cot\theta$ equal to?
- Consider the following statements :
1. In a triangle ABC, if \(\sin A + \sin B + \sin C = \frac{3\sqrt{3}}{2}\), then the triangle can be equilateral.
2. In a triangle ABC, if \(\cos A + \cos B + \cos C = \frac{3}{2}\), then the triangle can be equilateral.
Which of the statements given above is/are correct? What is \((1 + \tan\alpha \tan\beta)^2 + (\tan\alpha - \tan\beta)^2\) equal to ?
Consider the following statements :
I. \(\tan 50^\circ - \cot 50^\circ\) is positive
II. \(\cot 25^\circ - \tan 25^\circ\) is negative
Which of the statements is/are correct?
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