Question

For $n\in\mathbb Z$, a set of values of $\theta$ satisfying \[ \sec\theta-1=(\sqrt2-1)\tan\theta \] is

• A. \[ 2n\pi-\frac{\pi}{4} \]
• B. \[ 2n\pi+\frac{\pi}{2} \]
• C. \[ (2n+1)\pi+\frac{\pi}{4} \]
• D. \[ 2n\pi+\frac{\pi}{4} \]

Hint:

Whenever $1-\cos\theta$ appears, convert it into a half-angle expression.

Answer 1

The Correct Option is D

Step 1: Concept
Express everything in terms of sine and cosine.

Step 2: Meaning
Using \[ \sec\theta=\frac1{\cos\theta}, \qquad \tan\theta=\frac{\sin\theta}{\cos\theta}, \] the equation becomes \[ \frac{1-\cos\theta}{\cos\theta} = (\sqrt2-1)\frac{\sin\theta}{\cos\theta}. \]

Step 3: Analysis
Therefore \[ 1-\cos\theta=(\sqrt2-1)\sin\theta. \] Using \[ 1-\cos\theta = 2\sin^2\frac{\theta}{2}, \] and \[ \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, \] we get \[ \sin\frac{\theta}{2} = (\sqrt2-1)\cos\frac{\theta}{2}. \] Hence \[ \tan\frac{\theta}{2} = \sqrt2-1 = \tan\frac{\pi}{8}. \] Thus \[ \frac{\theta}{2} = n\pi+\frac{\pi}{8}. \]

Step 4: Conclusion
Therefore \[ \theta = 2n\pi+\frac{\pi}{4}. \]

Final Answer: (D)