The question asks to solve $\operatorname{arcsec}(-2) $. Options are a) π/3 b) −(π/3) c) 5π/3 d) 4π/3 e) 2π/3 So my thoughts are: this is quadrant 2 or 3; on the unit ci...

What is $\\def\\arcsec{\\operatorname{arcsec}}\\tan(\\arcsec(x))$ simplified and why? More specifically, I followed this reasoning, but apparently it is wrong: $\\tan ...

Dec 16, 2020 · However, because we know that $0 \leq y \leq \pi$ (and the range of $\text{arcsec}$ is precisely that), we can draw the above conclusion. This should be somewhat intuitive, since $\sec$ is simply the reciprocal function of $\cos$ .

May 20, 2018 · After using #y=x^2# and #2x*dx=dy# transforms, this integral became #1/2int arcsecy*dy# After using #z=arcsecy# , #y=secz# and #dy=secz*tanz*dz# transforms, it became

Jul 9, 2015 · First, let's find the derivative of #arcsec(x)# by applying the Chain Rule to the equation #sec(arcsec(x))=x# (and using the fact that #d/dx(sec(x))=sec(x)tan(x)#)

Feb 25, 2016 · My calculus book gives for the derivative of $\operatorname{arcsec}(x)$: $$\frac{d}{dx} \sec^{-1}x=\frac{1}{|x|\sqrt{x^2-1}}$$ Then it continues to state that: "Some authors prefer to define $\sec^{-1}$ as the inverse of the restriction of $\sec(x)$ to the separated intervals $[0,\pi/2)$ and $[\pi,3\pi/2)$ because this prevents the absolute ...

Sep 22, 2017 · 801 views around the world You can reuse this answer ...

May 15, 2014 · y = arcsec(1/x) Now, we have established that Arccos(x) = y = arcsec(1/x), Thus, arccos(x) = arcsec(1/x) And we have got a neat algebraic proof. Conversely, we can do the reverse by substituting cos for sec in the above equations to get arcsec(x) = arccos(1/x) Take a graphing calculator, you can substitute the values, and this is the image of ...

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Jun 15, 2017 · Next, the derivative of $\operatorname{arcsec}x$ is $\dfrac1{|x|\sqrt{x^2-1}}$. The absolute value is necessary in the denominator. In your case, this will proceed as follows: