Jul 25, 2016 · I changed it to cos: I know that sec(0)=1/cos(0) and that cos(0)=1 so: sec(0)=1

Oct 17, 2015 · cot0: undefined csc0: undefined sec0=1 0 is a special angle. cot, csc, and sec are actually just the reciprocals of tan, sin, and cos respectively.

Sep 23, 2016 · #tan^2x+3secx+3=0# #hArrsec^2x-1+3secx+3=0# or #sec^2x+3secx+2=0# or #sec^2x+secx+2secx+2=0# or #secx(secx+1)+2(secx+1)=0# or #(secx+2)(secx+1)=0# i.e. #secx=-2# or #secx=-1# If #secx=-2#, in the range #0^o,=x,360^o#, #x=120^o# or #x=240^o# and. if #secx=-1#, #x=180^o# Hence solution is #x=120^o# or #x=180^o# or #x=240^o#

May 3, 2018 · Suppose tan(θ) is undefined and sin(θ) < 0. Find the exact values of the 6 trigonometric functions of the angle θ?

Oct 8, 2016 · What is the average value of a function #f(t)=−4sec(t)tan(t)# on the interval #[0, pi/4]#? Calculus Applications of Definite Integrals The Average Value of a Function 1 Answer

Nov 16, 2017 · I got f'(0) = 0 We have: f(x) = 1/cosx = (cosx)^-1 By the chain rule, letting y = u^-1 and u = cosx, then y' = -1/u^2 and u' = -sinx. f'(x) = -1/u^2 * -sinx = -1/cos^2x * -sinx = tanxsecx We simplify evaluate x= 0 within the derivative. f'(0) = tan(0)sec(0) = 0 Hopefully this helps!

Jun 5, 2016 · What is the average value of a function #y=secx tanx# on the interval #[0,pi/3]#? Calculus Applications of Definite Integrals The Average Value of a Function 1 Answer

Nov 2, 2015 · Find 6 trig function values, knowing sec x = 6/5 and tan x < 0. sec x = 6/5 --> cos x = 5/6 --> arc x = +- 33^@56. Since tan x < 0, then x = -33^@56. sin (- 33.56) = - 0.55 tan (-33.56) = - 0.55/0.83 = - 0.66 cot (-33.56) = 1/(tan) = - 1.51 sec ( …

Nov 20, 2017 · There are three solutions to this equation: x_1 = \pi/3; \qquad x_2 = (5\pi)/3; and x_3 = \pi; Setting t=\secx, we get the following quadratic equation: t^2-t-2=0 which has as solutions, t_1 = 2; \qquad t_2 = -1; So we need to look for solutions to the equations - [1] \secx = 2 : \qquad \cosx = 1/2 \rightarrow x_1 = \pi/3; \quad x_2 = (5\pi)/3; [2] \secx = -1: \qquad \cosx=-1 \rightarrow x_3 ...

Feb 5, 2016 · That improper integral diverges (does not exist). Because secx is not defined at pi/2 which is less than 2, the integral is improper. Improper integral are probably not suitable for an "Introduction to Integration". In the early portion of a course on integration, the definite integral is often defined for function defined on some closed interval [a,b]. …