Q63 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Mathematics
If \(\cos \theta \neq 0\), and \(\sec \theta-1=(\sqrt{2}-1) \tan \theta\) then \(\theta=\)
- A\(n \pi+\frac{\pi}{8}, n \in Z\)
- Correct answerB\(2 n \pi+\frac{\pi}{4}\) or \(2 n \pi, n \in Z\)
- C\(2 n \pi+\frac{\pi}{8}, n \in Z\)
- D\(2 n \pi-\frac{\pi}{4}\) or \(2 n \pi, n \in Z\)
Solution
If \(\cos \theta \neq 0\) and \(\sec \theta-1=(\sqrt{2}-1) \tan \theta\)
\(\begin{array}{ll}
\Rightarrow \quad \frac{1-\cos \theta}{\cos \theta}=(\sqrt{2}-1) \frac{\sin \theta}{\cos \theta} \\
\Rightarrow \quad 2 \sin ^2 \frac{\theta}{2}=(\sqrt{2}-1) 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}
\end{array}\)
\(\Rightarrow \text{Either } \sin \frac{\theta}{2}=0 \text { or } \tan \frac{\theta}{2}=\sqrt{2}-1\)
\(\Rightarrow \text{Either } \frac{\theta}{2}=n \pi, n \in Z\)
\(\text{or } \frac{\theta}{2}=n \pi+\frac{\pi}{8}, n \in Z\)
\(\Rightarrow \quad \theta=2 n \pi+\frac{\pi}{4}\)
or \(2 n \pi, n \in Z\)
Hence, option (2) is correct.
Hence, the correct answer is option: 'B'.