x = tan-1(5) = 78.69 degrees
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant
x/2 = x/5 + 1 5x = 2x + 10 3x = 10 x = 3 1/3
x = 3.5
5/8 = tan(x) x = tan-1(5/8) = tan-1(0.625) = 0.558599 + k*pi radians or 32.00538 + k*180 degrees where k is an integer
When x = 3.806663, tan(e^x) = 1.
x = tan-1(5) = 78.69 degrees
cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
First: note 3 things about cot and tan, and note the given statement:cot = 1/tantan is cyclic with a period of π, that is tan(nπ + x) = tan(x)tan is an odd function, that is tan(-x) = -tan(x)tan(π/4) = 1Now apply them to the problem:cot(π - π/4) = 1/tan(π - π/4)= 1/tan(-π/4)= 1/-tan(π/4)= 1/-1 = -1Thus:cot(π - π/4) = -1.
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
tan(b) = x/sqrt(y^2-x^2)
It isn't, but here is a corrected equation that is: tan1 x tan 20 x tan 45 x tan 70 x tan 89 = 1 It will then work because tan(45 + x) = 1/tan(45 - x). Hence tan1 cancels when multiplied with tan89, and tan20 cancels with tan70. tan 45 equals 1, and so the whole expression cancels to 1. I think this effect was what the equation was intended to demonstrate.
sqrt(3sin(x)=cos(x)=0 // Square both sides3sin(x) + cos(x) = 0 // subtract cos(x) from both sides3sin(x) = -cos(x) // rearrangesin(x)/cos(x) = -1/3 //sin(x)/cos(x) = tan(x)tan(x) = -1/3x = tan^-1(-1/3) == -18,43484882 // tan^-1(inverse tan)
Let x = theta, since it's easier to type, and is essentially the same variable. Since tan^2(x)=tan(x), you know that tan(x) must either be 1 or zero for this statement to be true. So let tan(x)=0, and solve on your calculator by taking the inverse. Similarly for, tan(x)=1
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
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan 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