Question

If $\cos \theta \ne 0$ and $\sec \theta -1=(\sqrt{2}-1)\tan \theta$ then $\theta$ =

• A. $n\pi +\frac{\pi }{8},n\in Z$
• B. $2n\pi +\frac{\pi }{4}(or)2n\pi ,n\in Z$
• C. $2n\pi +\frac{\pi }{8},n\in Z$
• D. $2n\pi -\frac{\pi }{4}(or)2n\pi ,n\in Z$

Answer 1

Answer: (B)

If $\cos \theta \ne 0$ and $\sec \theta -1=(\sqrt{2}-1)\tan \theta$
$\Rightarrow \frac{1-\cos \theta }{\cos \theta }=(\sqrt{2}-1)\frac{\sin \theta }{\cos \theta }$
$\Rightarrow 2{\sin }^2\frac{\theta }{2}=(\sqrt{2}-1)2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$
$\Rightarrow$ Either $\sin \frac{\theta }{2}=0$ or $\tan \frac{\theta }{2}=\sqrt{2}-1$
$\Rightarrow$ Either $\frac{\theta }{2}=n\pi ,n\in Z$
or $\frac{\theta }{2}=n\pi +\frac{\pi }{8},n\in Z$
$\Rightarrow \theta =2n\pi +\frac{\pi }{4}$
or $2n\pi ,n\in Z$