Generating...                               s20quiz10_n5

  1. Suppose that $\frac{3\pi}{2} < t < 2\pi$, and that $\displaystyle \csc t=-2$ . Find the values of the trigonometric functions $\sin(t)$, $\cos(t)$, and $\tan(t)$.

    $\displaystyle \left[ \sin t=-{{1}\over{3}} , \cos t=-{{1}\over{2}} , \tan t={{2 }\over{3}} \right] $

    $\displaystyle \left[ \sin t=-{{1}\over{2}} , \cos t={{\sqrt{3}}\over{2}} , \tan t= -{{1}\over{\sqrt{3}}} \right] $

    $\displaystyle \left[ \sin t={{\sqrt{3}}\over{2}} , \cos t=-{{1}\over{2}} , \tan t= -\sqrt{3} \right] $

    $\displaystyle \left[ \sin t=-{{1}\over{2}} , \cos t=-{{1}\over{3}} , \tan t={{3 }\over{2}} \right] $

  2. Choose the correct trigonometric identity.

    $\displaystyle -\sin x-\cos x\,\cot x=2\,\sin x-\csc x$

    $\displaystyle -\sin x-\cos x\,\cot x=-\csc x$

    $\displaystyle -\sin x-\cos x\,\cot x={{2\,\csc x}\over{\left(\sec x\right)^2}}- \csc x$

    $\displaystyle -\sin x-\cos x\,\cot x=\csc x$

  3. Find the derivative $f'(x)$ for $f(x) =\displaystyle \sec x+3\,\cot x $ .

    $f'(x) =\displaystyle \sec x\,\tan x-3\,\left(\csc x\right)^2 $ $f'(x) =\displaystyle 3\,\tan x+\csc x $ $f'(x) =\displaystyle 3\,\left(\sec x\right)^2-\cot x\,\csc x $ $f'(x) =\displaystyle \cot x\,\csc x-3\,\left(\sec x\right)^2 $ $f'(x) =\displaystyle 3\,\left(\csc x\right)^2-\sec x\,\tan x $

  4. Find the derivative $f'(x)$ for $f(x) =\displaystyle {{\cos x}\over{x}} $ .

    $f'(x) =\displaystyle -{{\sin x-x\,\cos x}\over{x^2}} $ $f'(x) =\displaystyle -{{x\,\sin x+\cos x}\over{x^2}} $ $f'(x) =\displaystyle {{x\,\sin x+\cos x}\over{x^2}} $ $f'(x) =\displaystyle {{\sin x-x\,\cos x}\over{x^2}} $

  5. Choose the correct trigonometric identity.

    $\displaystyle \left(\csc t\right)^2+\left(\cot t\right)^2={{\cos ^2t+1}\over{ \sin ^2t}}$

    $\displaystyle \left(\csc t\right)^2+\left(\cot t\right)^2=1$

    $\displaystyle \left(\csc t\right)^2+\left(\cot t\right)^2=-{{\cos ^2t+1}\over{ \sin ^2t}}$

    $\displaystyle \left(\csc t\right)^2+\left(\cot t\right)^2=-1$

  6. Find the derivative $f'(x)$ for $f(x) =\displaystyle e^{x}\,\sec x $ .

    $f'(x) =\displaystyle -e^{x}\,\sec x\,\left(\tan x+1\right) $ $f'(x) =\displaystyle e^{x}\,\left(\cot x-1\right)\,\csc x $ $f'(x) =\displaystyle -e^{x}\,\left(\cot x-1\right)\,\csc x $ $f'(x) =\displaystyle e^{x}\,\sec x\,\left(\tan x+1\right) $

  7. Suppose that $\displaystyle g\left( {{4\,\pi}\over{3}} \right) = 1 $ and $\displaystyle g'\left( {{4\,\pi}\over{3}} \right) = 2 $. Then find the derivative $\displaystyle f'\left( {{4\,\pi}\over{3}} \right)$ for $f(x) = \displaystyle g\left(x\right)\,\sin x $ .

    $\displaystyle -\sqrt{3}-{{1}\over{2}}$ 2 $\displaystyle \sqrt{3}-{{1}\over{2}}$ $\displaystyle -{{\sqrt{3}}\over{2}}$ $\displaystyle {{1}\over{2}}-\sqrt{3}$

  8. Suppose that $\frac{\pi}{2} < t < \pi$, and that $\displaystyle \csc t=5$ . Find the values of the trigonometric functions $\sin(t)$, $\cos(t)$, and $\tan(t)$.

    $\displaystyle \left[ \sin t={{1}\over{5}} , \cos t=-{{2\,\sqrt{6}}\over{5}} , \tan t=-{{1}\over{2\,\sqrt{6}}} \right] $

    $\displaystyle \left[ \sin t=-{{2\,\sqrt{6}}\over{5}} , \cos t={{1}\over{5}} , \tan t=-2\,\sqrt{6} \right] $

    $\displaystyle \left[ \sin t={{2\,\sqrt{6}}\over{5}} , \cos t={{1}\over{5}} , \tan t=2\,\sqrt{6} \right] $

    $\displaystyle \left[ \sin t={{1}\over{5}} , \cos t={{2\,\sqrt{6}}\over{5}} , \tan t={{1}\over{2\,\sqrt{6}}} \right] $

  9. Choose the correct trigonometric identity.

    $\displaystyle \left(1-\sin ^2t\right)\,\left(\tan ^2t+1\right)=2\,\cos ^2t-1$

    $\displaystyle \left(1-\sin ^2t\right)\,\left(\tan ^2t+1\right)={{\sin ^2t+1 }\over{\cos ^2t}}$

    $\displaystyle \left(1-\sin ^2t\right)\,\left(\tan ^2t+1\right)=1$

    $\displaystyle \left(1-\sin ^2t\right)\,\left(\tan ^2t+1\right)=-{{2\,\sin ^4t+ \sin ^2t-1}\over{\cos ^2t}}$

  10. Choose the correct trigonometric identity.

    $\displaystyle -{{\sin \left(2\,t\right)}\over{\cos \left(2\,t\right)+1}}=\cot t$

    $\displaystyle -{{\sin \left(2\,t\right)}\over{\cos \left(2\,t\right)+1}}=-\tan t$

    $\displaystyle -{{\sin \left(2\,t\right)}\over{\cos \left(2\,t\right)+1}}=-\cot t$

    $\displaystyle -{{\sin \left(2\,t\right)}\over{\cos \left(2\,t\right)+1}}=\tan t$


Department of Mathematics
Last modified: 2026-03-28