Ex 34, 8 Find the general solution of the equation sec2 2x = 1 – tan 2x sec2 2x = 1 – tan 2x 1 tan2 2x = 1 – tan2x tan2 2x tan2x = 1 – 1 tan2 2x tan2x = 0 tan 2x (tan2x 1) = 0 Hence We know that sec2 x = 1 tan2 x So, sec2 2x = 1 tan2 2x tan 2x = 0 taA)cot x b)csc x c)tan x d)sec x tan x Please help me ( Math What is a simplified form of the expression sec^2x1/(sinx)(secx)? LH S = tan2x secx 1 1 = sec2x −1 secx 1 1 = (secx 1)(secx − 1) secx 1 1 *You can now get rid of (secx1) at the top and bottom of the fraction When the numerator and denominator of a fraction are both the same, providing they aren't both zeros, what you get is 1 = secx − 1 1 = secx = RH S
Find The Derivative Of The Given Function Y Tan 2x 1 Cot 2x I Tried Converting The Original Function In Terms Of Sin And Cos But It Was Still Too Complicated To Be Called Simplified
