Suppose that $0 < t < \frac{\pi}{2}$ , and that $\displaystyle \csc t=2$ . Find the values of the trigonometric functions $\sin(t)$ , $\cos(t)$ , and $\tan(t)$ .
$\displaystyle \left[ \sin t=-{{1}\over{3}} , \cos t={{1}\over{2}} , \tan t=-{{2 }\over{3}} \right]$
$\displaystyle \left[ \sin t={{1}\over{2}} , \cos t={{\sqrt{3}}\over{2}} , \tan t= {{1}\over{\sqrt{3}}} \right]$
$\displaystyle \left[ \sin t={{\sqrt{3}}\over{2}} , \cos t={{1}\over{2}} , \tan t= \sqrt{3} \right]$
$\displaystyle \left[ \sin t={{1}\over{2}} , \cos t=-{{1}\over{3}} , \tan t=-{{3 }\over{2}} \right]$
Choose the correct trigonometric identity.
$\sin( t+{{\pi}\over{2}}) = -\sin t$
$\sin( t+{{\pi}\over{2}}) = \cos t$
$\sin( t+{{\pi}\over{2}}) = \sin t$
$\sin( t+{{\pi}\over{2}}) = -\cos t$
Choose the correct trigonometric identity.
$\displaystyle \left(1-\sin ^2t\right)\,\left(\tan ^2t+1\right)=-{{2\,\sin ^4t+ \sin ^2t-1}\over{\cos ^2t}}$
$\displaystyle \left(1-\sin ^2t\right)\,\left(\tan ^2t+1\right)=1$
$\displaystyle \left(1-\sin ^2t\right)\,\left(\tan ^2t+1\right)=2\,\cos ^2t-1$
$\displaystyle \left(1-\sin ^2t\right)\,\left(\tan ^2t+1\right)={{\sin ^2t+1 }\over{\cos ^2t}}$
Choose the correct trigonometric identity.
$\displaystyle \sin x-\cos x\,\cot x=-\csc x$
$\displaystyle \sin x-\cos x\,\cot x=\csc x$
$\displaystyle \sin x-\cos x\,\cot x=2\,\sin x-\csc x$
$\displaystyle \sin x-\cos x\,\cot x={{2\,\csc x}\over{\left(\sec x\right)^2}}- \csc x$
Choose the correct trigonometric identity.
$\displaystyle \sin x\,\tan x+\cos x={{2\,\sec x}\over{\left(\csc x\right)^2}}- \sec x$
$\displaystyle \sin x\,\tan x+\cos x=\sec x$
$\displaystyle \sin x\,\tan x+\cos x=-\sec x$
$\displaystyle \sin x\,\tan x+\cos x=2\,\cos x-\sec x$
Find the derivative for $f(x) =\displaystyle {{\tan x}\over{x^2}}$ .
$f'(x) =\displaystyle {{x\,\left(\csc x\right)^2+2\,\cot x}\over{x^3}}$ $f'(x) =\displaystyle {{2\,\tan x-x\,\left(\sec x\right)^2}\over{x^3}}$ $f'(x) =\displaystyle -{{x\,\left(\csc x\right)^2+2\,\cot x}\over{x^3}}$ $f'(x) =\displaystyle -{{2\,\tan x-x\,\left(\sec x\right)^2}\over{x^3}}$
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 $\tan(t)$ .
$\displaystyle \left[ \sin t={{1}\over{5}} , \cos t=-{{2\,\sqrt{6}}\over{5}} , \tan t=-{{1}\over{2\,\sqrt{6}}} \right]$
$\displaystyle \left[ \sin t={{1}\over{5}} , \cos t={{2\,\sqrt{6}}\over{5}} , \tan t={{1}\over{2\,\sqrt{6}}} \right]$
$\displaystyle \left[ \sin t=-{{2\,\sqrt{6}}\over{5}} , \cos t={{1}\over{5}} , \tan t=-2\,\sqrt{6} \right]$
$\displaystyle \left[ \sin t={{2\,\sqrt{6}}\over{5}} , \cos t={{1}\over{5}} , \tan t=2\,\sqrt{6} \right]$
Suppose that $\displaystyle g\left( {{2\,\pi}\over{3}} \right) = -3$ and $\displaystyle g'\left( {{2\,\pi}\over{3}} \right) = -2$ . Then find the derivative $\displaystyle f'\left( {{2\,\pi}\over{3}} \right)$ for $f(x) = \displaystyle g\left(x\right)\,\cot x$ .
$\displaystyle \sqrt{3}$ $\displaystyle {{2}\over{\sqrt{3}}}-4$ $\displaystyle {{2}\over{\sqrt{3}}}+4$ −2 $\displaystyle 4-{{2}\over{\sqrt{3}}}$
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 $\tan(t)$ .
$\displaystyle \left[ \sin t=-{{1}\over{2}} , \cos t={{\sqrt{3}}\over{2}} , \tan t= -{{1}\over{\sqrt{3}}} \right]$
$\displaystyle \left[ \sin t=-{{1}\over{3}} , \cos t=-{{1}\over{2}} , \tan t={{2 }\over{3}} \right]$
$\displaystyle \left[ \sin t=-{{1}\over{2}} , \cos t=-{{1}\over{3}} , \tan t={{3 }\over{2}} \right]$
$\displaystyle \left[ \sin t={{\sqrt{3}}\over{2}} , \cos t=-{{1}\over{2}} , \tan t= -\sqrt{3} \right]$
Find the derivative for $f(x) =\displaystyle e^{x}\,\cot x$ .
$f'(x) =\displaystyle -e^{x}\,\left(\tan x+\left(\sec x\right)^2\right)$ $f'(x) =\displaystyle e^{x}\,\left(\left(\csc x\right)^2-\cot x\right)$ $f'(x) =\displaystyle -e^{x}\,\left(\left(\csc x\right)^2-\cot x\right)$ $f'(x) =\displaystyle e^{x}\,\left(\tan x+\left(\sec x\right)^2\right)$

Department of Mathematics
Last modified: 2025-06-13