Question

If $\cot \theta - \tan \theta = \sec \theta$ , then $\theta$ is equal to

• A. $n\pi+(-1)^n\frac{\pi}{6}$
• B. $n\pi+\frac{\pi}{2}$
• C. $2n\pi+3\frac{\pi}{2}$
• D. none of these.

Answer 1

Answer: (A)

$ \cot \theta - \tan \theta = \sec \theta$
$\Rightarrow $ $\frac{\cos \, \theta }{\sin \, \theta } - \frac{\sin \, \theta }{\cos \, \theta } = \frac{1}{\cos \, \theta } $
$\, \, \, \, \, \, \, \, \, \, \, \, [\cos \theta \, \neq 0, \sin \theta \, \neq 0] $
$\Rightarrow $ $\, \, \cos^2 \, \theta -\sin^2 \,\theta = \sin\, \theta $
$\Rightarrow $ $\, \, \, \, \, \, \, \, \, 1 - 2 \, \sin^2 \theta = \sin \, \theta$
$\Rightarrow $ $\, \, \, \, \, 2 \, \sin^2 \, \theta + \sin \, \theta- 1 = 0$
$\Rightarrow $ $(2 \, \sin \,\theta - 1) (\sin\, \theta + 1) = 0$
$\Rightarrow $ $\, \, \, \sin \, \theta = \frac{1}{2} = \sin \frac{\pi}{6} $ $[ \because\, \sin \, \theta \, \neq - 1 ] $
$\Rightarrow $ $\, \, \, \, \, \, \theta = n \pi + (-1)^n \, \frac{ \pi}{6}$