Question

If \( \cos\theta + \sin\theta = \sqrt{2}\cos\theta \) and \( 0<\theta<\frac{\pi}{2} \), then \( \sec(2\theta) + \tan(2\theta) = \)

• A. \( \cot\theta \)
• B. \( \tan\theta \)
• C. \( \cos\theta \)
• D. \( \sin\theta \)

Hint:

The expression \( \sec(x) + \tan(x) \) simplifies to \( \tan(\frac{x}{2} + \frac{\pi}{4}) \). Alternatively, the identity \( \frac{1+\sin(2\theta)}{\cos(2\theta)} = \frac{1+\tan\theta}{1-\tan\theta} \) is a very useful shortcut for this type of problem.

Answer 1

The Correct Option is A

First, simplify the given equation to find \( \tan\theta \).
\( \sin\theta = \sqrt{2}\cos\theta - \cos\theta = (\sqrt{2}-1)\cos\theta \).
Dividing both sides by \( \cos\theta \) (which is non-zero as \( 0<\theta<\pi/2 \)), we get:
\( \tan\theta = \sqrt{2}-1 \).
Now, simplify the expression we need to find.
\( \sec(2\theta) + \tan(2\theta) = \frac{1}{\cos(2\theta)} + \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{1+\sin(2\theta)}{\cos(2\theta)} \).
Using trigonometric identities: \( 1 = \cos^2\theta + \sin^2\theta \), \( \sin(2\theta) = 2\sin\theta\cos\theta \), and \( \cos(2\theta) = \cos^2\theta - \sin^2\theta \).
Numerator: \( 1+\sin(2\theta) = \cos^2\theta + \sin^2\theta + 2\sin\theta\cos\theta = (\cos\theta + \sin\theta)^2 \).
Denominator: \( \cos(2\theta) = \cos^2\theta - \sin^2\theta = (\cos\theta - \sin\theta)(\cos\theta + \sin\theta) \).
The expression becomes \( \frac{(\cos\theta + \sin\theta)^2}{(\cos\theta - \sin\theta)(\cos\theta + \sin\theta)} = \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \).
Divide the numerator and denominator by \( \cos\theta \): \( \frac{1 + \tan\theta}{1 - \tan\theta} \).
Substitute \( \tan\theta = \sqrt{2}-1 \): \( \frac{1 + (\sqrt{2}-1)}{1 - (\sqrt{2}-1)} = \frac{\sqrt{2}}{2-\sqrt{2}} \).
Rationalize the denominator: \( \frac{\sqrt{2}}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}} = \frac{2\sqrt{2}+2}{4-2} = \frac{2(\sqrt{2}+1)}{2} = \sqrt{2}+1 \).
Now, let's evaluate the correct option, \( \cot\theta \).
\( \cot\theta = \frac{1}{\tan\theta} = \frac{1}{\sqrt{2}-1} \).
Rationalize: \( \frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1 \).
Both expressions evaluate to the same value, so the answer is correct.