May 23, 2018 · tanθ + secθ = 1....[1] Now We know sec^2theta-tan^2theta=1 =>sectheta-tantheta=1/(sectheta+tantheta)=1/1 =>sectheta-tantheta=1.....[2]

Aug 29, 2014 · The derivative of #y = tan(x)sec(x)# with respect to #x# is #(dy)/(dx) = sec^3(x) + sec(x)tan^2(x)#. To perform this differentiation, we will need to use the Product rule, which states that given the product of two functions, #u(x)v(x)#, here represented as #f(x) = u(x)v(x)#...

Feb 20, 2018 · sec x/tan x (1/cosx )/(sinx/cosx) 1/cancelcosx xx cancelcosx/sinx 1/sinx = cscx

Sep 11, 2014 · By Product Rule, we can find y'=secx(1+2tan^2x). Let us look at some details. y=secxtanx By Product Rule, y'=secxtanx cdot tanx+secx cdot sec^2x by factoring out sec x, =secx(tan^2x+sec^2x) by sec^2x=1+tan^2x, =secx(1+2tan^2x)

Sep 25, 2016 · One of the Pythagorean trigonometric identities is . #tan^2 x + 1 = sec^2 x# Which also gets us #sec^2 x - tan^2 x = 1#.

Jun 14, 2018 · -cscx+C Let's rewrite the integrand: intsecx/tan^2xdx=int(1/cosx)/(sin^2x/cos^2x)dx=intcosx/sin^2x Here, recognize that intcosx/sin^2xdx=intcosx/sinx1/sinxdx ...

Apr 27, 2018 · y/x 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.

Jul 30, 2015 · Simplify: (1 + tan x)/sec x Ans: cos x + sin x (1 + sin x/cos x)/(1/cos x) = ((cos x + sin x)/cos x)(cos x/1) = cos x + sin x

Nov 27, 2016 · If #sec theta + tan theta = P#, then what is #cos theta# equal to? Trigonometry Right Triangles Relating Trigonometric Functions. 1 Answer

Apr 25, 2018 · See below Apply trig identity: 1+tan^2x=sec^2x So: 2sec^2x + tan^2x-3=0 2(1+tan^2x)+tan^2x-3=0 2+2tan^2x+tan^2x-3=0 3tan^2x-1=0 tan^2x=1/3 tanx= +-sqrt3/3 x=pi/6+-pin x= (5pi)/6+-pin Where n is an element of all real numbers graph{2(secx)^2+(tanx)^2-3 [-10, 10, …