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tan x + (tan x)(sec 2x) = tan 2x work dependently on the left side
tan x + (tan x)(sec 2x); factor out tan x
= tan x(1 + sec 2x); sec 2x = 1/cos 2x
= tan x(1 + 1/cos 2x); LCD = cos 2x
= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x
= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]
= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x
= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x
= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x
= sin 2x/cos 2x
= tan 2x
What else can I help you with?
What is the integral sec x?
ln|sec x + tan x| + C.
What is the derivative of secxtanx?
d/dx(uv)=u*dv/dx+v*du/dxd/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]-The derivative of tanx is:d/dx(tan u)=[sec(u)]2*d/dx(u)d/dx(tan x)=[sec(x)]2*d/dx(x)d/dx(tan x)=[sec(x)]2*(1)d/dx(tan x)=(sec(x))2=sec2(x)-The derivative of secx is:d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)d/dx(sec x)=[sec(x)tan(x)]*(1)d/dx(sec x)=sec(x)tan(x)d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)
Sec x times sin x divided by tan x?
1 (sec x)(sin x /tan x = (1/cos x)(sin x)/tan x = (sin x/cos x)/tan x) = tan x/tan x = 1
How do you prove sin x tan x equals cos x?
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
What is integration of secx tanx?
Will try integration by parts. uv - int[v du] u = sec(x)----------------du = sec(x) tan(x) dv = tan(x)---------------v = ln[sec(x)] sec(x) ln[sex(x)] - int[lnsec(x) dx] = sec(x) ln[sec(x)] - xlnsec(x) - x + C ===========================
Sec squared theta plus tan squared theta equals to 13 12 and sec raised to 4 theta minus tan raised to 4 theta equals to how much?
It also equals 13 12.
How do you prove tan x plus sin x equals 2 tan x?
This would be a real bear to prove, mainly because it's not true.
What is the derivative of y equals sec x?
sec(x)tan(x)
Csc divided by cot squared equals tan multiplied by sec?
Yes.
Tan 9-tan 27-tan 63 plus tan81 equals 4?
4
Sec - cos equals tansin?
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
Tan3x plus tanx equals 2tan2x?
Tan
Can you help me prove math trig identities?
these are the identities i need sinΘcosΘ=cosΘ sec^4Θ-tan^4Θ=sec²Θcsc²Θ (1+sec²Θ)/(1-secΘ)=(cosΘ-1)/(cosΘ)
How do you solve Sin x sec x equals tan x?
Cos x = 1 / Sec x so 1 / Cos x = Sec x Then Tan x = Sin x / Cos x = Sin x * (1 / Cos x) = Sin x * Sec x
How do you identify sec x sin x equals tan x?
Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.
What is theta when tantheta plus sectheta equals 1 please show work?
copy this and paste in your browsers address window http://www.wolframalpha.com/input/?i=tan+theta+%2B+sec+theta+%3D1
What is the drevative of f of x equals tan redical x power2 plus 1?
let u = sqrt(x^2 + 1) d/dx ( tan u ) = du/dx sec^2 u du/dx = x/sqrt(x^2 + 1) d/dx (tan u) = x/sqrt(x^2 + 1) times sec^2 (sqrt(x^2 + !) ( I think)
