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If $\sec \alpha=\frac{13}{5}$ where $270^{\circ} < \alpha < 360^{\circ}$ then what is $\sin \alpha$ equal
to ?
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to ?
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Verified Answer
The correct answer is:
$-\frac{12}{13}$
We have sec $\alpha=\frac{13}{5}$
Since $\frac{3 \pi}{2} < \alpha < 2 \pi$
$\therefore \sin \alpha < 0$
Now, $\sin \alpha=-\sqrt{1-\frac{1}{\sec ^{2} \alpha}}$
$=-\sqrt{1-\frac{25}{169}}=-\sqrt{\frac{144}{169}}=-\frac{12}{13}$
133. (b) $\tan \left(-585^{\circ}\right)=-\tan 585^{\circ}$
$=-\tan \left[540^{\circ}+45^{\circ}\right]=-\tan \left[3 \pi+\frac{\pi}{4}\right]$
$=-\left[\frac{\tan 3 \pi+\tan \frac{\pi}{4}}{1-\tan 3 \pi \tan \frac{\pi}{4}}\right]=-\left[\frac{0+1}{1-0 \times 1}\right]=-1$
Since $\frac{3 \pi}{2} < \alpha < 2 \pi$
$\therefore \sin \alpha < 0$
Now, $\sin \alpha=-\sqrt{1-\frac{1}{\sec ^{2} \alpha}}$
$=-\sqrt{1-\frac{25}{169}}=-\sqrt{\frac{144}{169}}=-\frac{12}{13}$
133. (b) $\tan \left(-585^{\circ}\right)=-\tan 585^{\circ}$
$=-\tan \left[540^{\circ}+45^{\circ}\right]=-\tan \left[3 \pi+\frac{\pi}{4}\right]$
$=-\left[\frac{\tan 3 \pi+\tan \frac{\pi}{4}}{1-\tan 3 \pi \tan \frac{\pi}{4}}\right]=-\left[\frac{0+1}{1-0 \times 1}\right]=-1$
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