precal
posted by Anonymous .
sec^4xtan^4x=1+2tan^2x

sec^4 = (tan^2+1)^2 = tan^4 + 2tan^2 + 1
subtract and voila.
Respond to this Question
Similar Questions

Integration
Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct? 
calculus
find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x … 
Math  Trigonometry
Verify the following: 1. cos x/(1sinx)= sec x + tan x 2. (tanx+1)^2=sec^2x + 2tan x 3. csc x = )cot x + tan x)/sec x 4. sin2x  cot x = cotxcos2x 
Trig.
sec^2xcotxcotx=tanx (1/cos)^2 times (1/tan)(1/tan)=tan (1/cos^2) times (2/tan)=tan (2/cos^2tan)times tan=tan(tan) sq. root of (2/cos^2)= sq. root of (tan^2) sq. root of (2i)/cos=tan I'm not sure if I did this right. If I didn't, … 
Calculus
Evaluate the indefinite integral integral sec(t/2) dt= a)ln sec t +tan t +C b)ln sec (t/2) +tan (t/2) +C c)2tan^2 (t/2)+C d)2ln cos(t/2) +C e)2ln sec (t/2)+tan (t/2) +C 
trig
prove: (tan x + sec x)^2 = 2sec^2 x + 2tan x sec x  1 
trig
prove: (tan x + sec x)^2 = 2sec^2 x + 2tan x sec x  1 
calculus (check my work please)
Not sure if it is right, I have check with the answer in the book and a few integral calculators but they seem to get a different answer ∫ sec^3(x)tan^3(x) dx ∫ sec^3(x)tan(x)(sec^2(x)1) dx ∫ tan(x)sec(x)[sec^4(x)sec^2(x)] … 
Trigonometry
1.Solve tan^2x + tan x – 1 = 0 for the principal value(s) to two decimal places. 6.Prove that tan y cos^2 y + sin^2y/sin y = cos y + sin y 10.Prove that 1+tanθ/1tanθ = sec^2θ+2tanθ/1tan^2θ 17.Prove that … 
Trigonometry desperate help, clueless girl here
2. solve cos 2x3sin x cos 2x=0 for the principal values to two decimal places. 3. solve tan^2 + tan x1= 0 for the principal values to two decimal places. 4. Prove that tan^2(x) 1 + cos^2(x) = tan^2(x) sin^2 (x). 5.Prove that tan(x) …