Question

If $\cos \theta +\sec \theta =-2,\theta \in [0,2\pi ],$ then ${\sin }^8\theta +{\cos }^8\theta$ equal to

• A. $-2$
• B. $1$
• C. $2^4$
• D. $2^5$

Answer 1

The correct option is B $1$

$\cos \theta +\sec \theta =-2$

$\cos \theta +\frac{1}{\cos \theta }=-2$

${\cos }^2\theta +2\cos \theta +1=0$

${(\cos \theta +1)}^2=0$

$\cos \theta =-1,\sin \theta =0$

${\sin }^8\theta +{\cos }^8\theta =0+{(-1)}^8=1$