sec2(x) = 1 + tan2(x)Leave the left side assec2(x) and for the proof, try to make the right side equal the left side.
We know tan = sin/cos
1 + tan2(x) = 1 + sin2(x)/cos2(x)
Rewrite the 1 in terms of cos2(x)
1 + sin2(x)/cos2(x) = cos2(x)/cos2(x) + sin2(x)/cos2(x)
Simplify, now that the denominators have the same terms
cos2(x)/cos2(x) + sin2(x)/cos2(x) = [ cos2(x) + sin2(x) ] / cos2(x)
Use the trig. identity: cos2(x) + sin2(x) = 1
[ cos2(x) + sin2(x) ] / cos2(x) = 1 / cos2(x)
Remember 1/cosine is equal to secant.\
1 / cos2(x) = sec2(x)
Recall, the left side the equation was: sec2(x)
The right side (we just solved for is): sec2(x)
sec2(x) = sec2(x)
Left side = right side, Q.E.D.
It also equals 13 12.
f(x)= tan2(x) f'(x)= 2tan(x)*sec2(x)
You use the identity sin2x + cos2x = 1 (to simplify the expression in parentheses), and convert all functions to sines and cosines. sec x tan x (1 - sin2x) = (1/cos x) (sin x / cos x) (cos2x) = (sin x / cos2x) cos2x = sin x
sec(x)=1/cos(x) - (hint: look at the third letter: sec->(1/)cos, cosec->(1/)sin, cot->(1/)tan)
tan(sqrtX) + C
It also equals 13 12.
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
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
A = Root (Q squared plus P squared) C = 90 + tan inverse P/Q ... I think lol just worked it out just now =w=
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
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
Let s = sin x; c = cos x. By definition, sec x = 1/cos x = 1/c; and tan x = (sin x) / (cos x) = s/c. We know, also, that s2 + c2 = 1. Then, dividing through by c2, we have, (s2/c2) + 1 = (1/c2), or (s/c)2 + 1 = (1/c)2; in other words, we have, tan2 x + 1 = sec2 x.