
How do I simplify sec x/tan x - Socratic
2018年2月20日 · sec x/tan x (1/cosx )/(sinx/cosx) 1/cancelcosx xx cancelcosx/sinx 1/sinx = cscx
How do you simplify Tan (sec^-1 (u))? | Socratic
2018年4月27日 · Trigonometric functions of an angle can be thought of as ratios of the sides of a triangle and its inverse functions are functions of a ratio of that triangle that return an angle. …
Sec (x)tan (x)=2.find the value of x, 0<x<360 - Socratic
2018年6月14日 · x = arcsin (-1/4 + sqrt (17)/4) First of all, apply the definitions: since sec (x) = 1/cos (x) and tan (x) = sin (x)/cos (x), we have sec (x)tan (x) = 1/cos (x)*sin (x)/cos (x) = sin …
How do you evaluate sec (tan^-1 sqrt (3))? | Socratic
2016年6月5日 · Set x=tan^-1 sqrt3=>tanx=sqrt3=>x=pi/3 Hence sec (tan^-1 sqrt3)=sec (pi/3)=2
sec A = 4 how do you find the value of cos A? - Socratic
2015年10月14日 · If tan A + sec A = 4 how do you find the value of cos A?
How do you prove sin theta sec theta- tan theta? | Socratic
2015年6月5日 · Assuming the minus sign should be an equal sign, so the question is to prove: color (white) ("XXXX")sin (theta)sec (theta) = tan (theta) sec (theta) = 1/cos (theta ...
What is Sec (tan^-1 (2))? - Socratic
2015年10月4日 · sec (arctan (2)) = sqrt (5) From the trigonometric identity sin^2 (theta) + cos^2 (theta) = 1, divide both sides by cos^2 (theta) sin^2 (theta)/cos^2 (theta) + cos^2 (theta)/cos^2 …
How do you find sin, cos, tan, sec, csc, and cot given (-4,-4)?
2018年3月2日 · See below. If we are given coordinates of the form (x,y), where x and y are negative then we are in the III quadrant. Since (-4,-4) are the sides of a right triangle, then the …
How do you solve sec (tan^-1x) = sqrt (1+x^2)? | Socratic
2018年6月9日 · Please go through Explanation. Recall that, tan^-1x=theta, x in RR iff tantheta=x, theta in (-pi/2,pi/2). Letting tan^-1x=theta, we have, sec (tan^-1x)=sectheta==sqrt (sec^2theta), …
If sec (theta) + tan (theta) = 1.5 then in which quadratant
2018年1月10日 · If sec(θ) + tan(θ) = 1.5 then in which quadratant does θ lies?1st quadrant... Hope it helps... Thank you...