For the following two (02) items: Let $\frac{1+\sin\theta}{\cos\theta} = p + \sqrt{p^2+1}$.
Trigonometric Solution: Finding \(\tan\theta\)
This problem involves simplifying a trigonometric expression and using fundamental identities to find the value of \(\tan\theta\) in terms of the variable \(p\). We are given the equation:
\(\frac{1+\sin\theta}{\cos\theta} = p + \sqrt{p^2+1}\)
Step 1: Simplify the Left-Hand Side Expression
We can rewrite the left-hand side (LHS) of the equation using basic trigonometric definitions:
- Recall that \(\sec\theta = \frac{1}{\cos\theta}\) and \(\tan\theta = \frac{\sin\theta}{\cos\theta}\).
Splitting the fraction:
\(\frac{1+\sin\theta}{\cos\theta} = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}\)
Substituting the identities:
\(\sec\theta + \tan\theta\)
So, the given equation can be expressed as:
\(\sec\theta + \tan\theta = p + \sqrt{p^2+1}\)
Step 2: Utilize the Pythagorean Identity
We know the fundamental Pythagorean identity involving secant and tangent:
\(\sec^2\theta - \tan^2\theta = 1\)
This identity can be factored as a difference of squares:
\((\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1\)
From this, we can express \((\sec\theta - \tan\theta)\) in terms of \((\sec\theta + \tan\theta)\):
\(\sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta}\)
Step 3: Substitute and Solve for \((\sec\theta - \tan\theta)\)
Using the result from Step 1, substitute the value of \((\sec\theta + \tan\theta)\):
\(\sec\theta - \tan\theta = \frac{1}{p + \sqrt{p^2+1}}\)
To simplify this expression, we rationalize the denominator by multiplying the numerator and denominator by the conjugate (\(p - \sqrt{p^2+1}\)):
\(\sec\theta - \tan\theta = \frac{1}{p + \sqrt{p^2+1}} \times \frac{p - \sqrt{p^2+1}}{p - \sqrt{p^2+1}}\)
\(\sec\theta - \tan\theta = \frac{p - \sqrt{p^2+1}}{p^2 - (\sqrt{p^2+1})^2}\)
\(\sec\theta - \tan\theta = \frac{p - \sqrt{p^2+1}}{p^2 - (p^2+1)}\)
\(\sec\theta - \tan\theta = \frac{p - \sqrt{p^2+1}}{-1}\)
\(\sec\theta - \tan\theta = \sqrt{p^2+1} - p\)
Step 4: Solve the System of Equations for \(\tan\theta\)
Now we have a system of two linear equations with \(\sec\theta\) and \(\tan\theta\):
- \(\sec\theta + \tan\theta = p + \sqrt{p^2+1}\)
- \(\sec\theta - \tan\theta = \sqrt{p^2+1} - p\)
To find \(\tan\theta\), subtract the second equation from the first equation:
\((\sec\theta + \tan\theta) - (\sec\theta - \tan\theta) = (p + \sqrt{p^2+1}) - (\sqrt{p^2+1} - p)\)
\(\sec\theta + \tan\theta - \sec\theta + \tan\theta = p + \sqrt{p^2+1} - \sqrt{p^2+1} + p\)
Combine like terms:
\(2\tan\theta = 2p\)
Divide by 2:
\(\tan\theta = p\)
Conclusion
By simplifying the given trigonometric expression and applying the Pythagorean identity, we found that \(\tan\theta\) is equal to \(p\).
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 \(\sec\theta\) equal to?
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- 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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