Generating...                               s20quiz10_n27

  1. Choose the correct trigonometric identity.

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

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

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

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

  2. Choose the correct trigonometric identity.

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

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

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

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

  3. Find the derivative $f'(x)$ for $f(x) =\displaystyle \csc x\,\tan x $ .

    $f'(x) =\displaystyle \sec x\,\tan x $ $f'(x) =\displaystyle -\sec x\,\tan x $ $f'(x) =\displaystyle \cot x\,\csc x $ $f'(x) =\displaystyle \csc x $ $f'(x) =\displaystyle -\cot x\,\csc x $

  4. Suppose that $\displaystyle g\left( {{5\,\pi}\over{3}} \right) = -1 $ and $\displaystyle g'\left( {{5\,\pi}\over{3}} \right) = 2 $. Then find the derivative $\displaystyle f'\left( {{5\,\pi}\over{3}} \right)$ for $f(x) = \displaystyle g\left(x\right)\,\tan x $ .

    $\displaystyle 4-2\,\sqrt{3}$ $\displaystyle \sqrt{3}$ $\displaystyle 2\,\sqrt{3}-4$ 2 $\displaystyle -2\,\sqrt{3}-4$

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

    $f'(x) =\displaystyle -e^{x}\,\left(\sin x+\cos x\right) $ $f'(x) =\displaystyle e^{x}\,\left(\sin x+\cos x\right) $ $f'(x) =\displaystyle -e^{x}\,\left(\sin x-\cos x\right) $ $f'(x) =\displaystyle e^{x}\,\left(\sin x-\cos x\right) $

  6. Find the derivative $f'(x)$ for $f(x) =\displaystyle {{\cot x}\over{x^3}} $ .

    $f'(x) =\displaystyle {{3\,\tan x-x\,\left(\sec x\right)^2}\over{x^4}} $ $f'(x) =\displaystyle -{{x\,\left(\csc x\right)^2+3\,\cot x}\over{x^4}} $ $f'(x) =\displaystyle -{{3\,\tan x-x\,\left(\sec x\right)^2}\over{x^4}} $ $f'(x) =\displaystyle {{x\,\left(\csc x\right)^2+3\,\cot x}\over{x^4}} $

  7. Choose the correct trigonometric identity.

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

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

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

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

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

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

  9. Choose the correct trigonometric identity.

    $\displaystyle \sin x\,\tan x-\cos x=2\,\cos x-\sec x$

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

    $\displaystyle \sin x\,\tan x-\cos x=\sec x$

    $\displaystyle \sin x\,\tan x-\cos x=-\sec x$

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

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

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

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

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


Department of Mathematics
Last modified: 2026-07-16