Question

$Ifcos\theta -tan\theta =sec\theta ,then,\theta isequalto$

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

Answer 1

The correct option is B

$n\pi +(-1{)}^n\frac{\pi }{6},n\in Z$

$Givenequation:cos\theta -tan\theta =sec\theta \Rightarrow \frac{cos\theta }{sin\theta }-\frac{sin\theta }{cos\theta }=\frac{1}{cos\theta }\Rightarrow \frac{co{s}^2\theta -si{n}^2\theta }{sin\theta cos\theta }=\frac{1}{cos\theta }\Rightarrow co{s}^2\theta -si{n}^2\theta =sin\theta \Rightarrow (1-si{n}^2\theta )-si{n}^2\theta =sin\theta \Rightarrow 1-2si{n}^2\theta +sin\theta -1=0\Rightarrow 2si{n}^2\theta +sin\theta -1=0\Rightarrow 2si{n}^2\theta +2sin\theta -sin\theta -1=0\Rightarrow 2sin\theta (sin\theta +1)-1(sin\theta +1)=0\Rightarrow (sin\theta +1)(2sin\theta -1)=0\Rightarrow sin\theta +1=0or2sin\theta -1=0\Rightarrow sin\theta =-1orsin\theta =\frac{1}{2}Now,sin\theta =-1\Rightarrow sin\theta =sin\frac{3\pi }{2}\Rightarrow \theta =m\pi +(-1{)}^m\frac{3\pi }{2},m\in ZAnd,sin\theta =\frac{1}{2}\Rightarrow sin\theta =sin\frac{\pi }{6}\Rightarrow \theta =n\pi +(-1{)}^n\frac{\pi }{6}n\in Z\therefore q=n\pi +(-1{)}^n\frac{\pi }{6},n\in Z$