Question

If $\sin \theta \sec \theta =-1$ and $\theta$ lies in the second quadrant, find $\sin \theta$ and $\sec \theta$.

• A. $\sin \theta =\frac{1}{\sqrt{2}},\sec \theta =-\sqrt{2}$
• B. $\sin \theta =-\frac{1}{\sqrt{2}},\sec \theta =\sqrt{2}$
• C. $\sin \theta =\sqrt{2},\sec \theta =\frac{1}{\sqrt{2}}$
• D. $\sin \theta =-\sqrt{2},\sec \theta =\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A)

$\sin \theta =\frac{1}{\sqrt{2}},\sec \theta =-\sqrt{2}$

Given: $\sin \theta \sec \theta =-1$

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

$\Rightarrow {\sin }^2\theta ={\cos }^2\theta$

$\Rightarrow {\sin }^2\theta =1-{\sin }^2\theta$

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

$\Rightarrow \sin \theta =\pm \frac{1}{\sqrt{2}}$

As $\theta$ lies in second quadrant ;

$\Rightarrow \theta =\frac{3\pi }{4}$

$\Rightarrow \sin \theta =\sin \frac{3\pi }{4}=\frac{1}{\sqrt{2}}$

$\Rightarrow \sec \theta =\sec \frac{3\pi }{4}=-\sqrt{2}$