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
It also equals 13 12.
This would be a real bear to prove, mainly because it's not true.
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.
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
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
Yes, it is.
sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
copy this and paste in your browsers address window http://www.wolframalpha.com/input/?i=tan+theta+%2B+sec+theta+%3D1
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
It is NOT equal. Try calculating tan x, and tan 6x, for a few values of "x", on your scientific calculator. Perhaps you are supposed to solve an equation, and see FOR WHAT values of "x" the two are equal?
these are the identities i need sinΘcosΘ=cosΘ sec^4Θ-tan^4Θ=sec²Θcsc²Θ (1+sec²Θ)/(1-secΘ)=(cosΘ-1)/(cosΘ)
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
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
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)
The derivative of sec(x) is sec(x) tan(x).
tan(x)*csc(x) = sec(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
integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C
ln|sec x + tan x| + C.
Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.
tan(23) = 1.58815308