Question

If $\cot \theta -\tan \theta =\sec \theta$, then $\theta$ =

• A. $n\pi +{(-1)}^n\pi /6$
• B. $n\pi +\pi /2$
• C. $2n\pi +3\pi /2$
• D. $Noneofthese$

Answer 1

The correct option is A $n\pi +{(-1)}^n\pi /6$

$cot\theta -tan\theta =sec\theta$

$\frac{cos\theta }{sin\theta }-\frac{sin\theta }{cos\theta }=\frac{1}{cos\theta }$

$\frac{co{s}^2\theta -si{n}^2\theta }{sin\theta cos\theta }=\frac{1}{cos\theta }$

$cos\theta (co{s}^2\theta -si{n}^2\theta )-sin\theta cos\theta =0$

$cos\theta =0$ or $co{s}^2\theta -si{n}^2\theta -sin\theta cos\theta =0$

$\Rightarrow co{s}^2\theta -si{n}^2\theta =sin\theta$

$1-2si{n}^2\theta =sin\theta$

$2si{n}^2\theta +sin\theta -1=0$

$sin\theta =\frac{-1\pm \sqrt{1+8}}{4}=\frac{-1\pm \sqrt{9}}{4}$

$sin\theta =\frac{-1\pm 3}{4}=\frac{1}{2}$ or $1$

As $\theta \ne \frac{\pi }{2}$ so,

$sin\theta =\frac{1}{2}$

$\theta =\frac{\pi }{6}+n\pi$ for $1^{st}$ and $2^{nd}$ Quadrant

$\theta =n\pi +(-1{)}^n\frac{\pi }{6}$