How do you simplify #tan^4 theta + 2tan^2 theta + 1 #?

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Answer 1 · 2015-10-18T20:00:24

#tan^4(theta) + 2tan^2(theta) + 1 = sec^4(theta)#

#tan^4(theta) + 2tan^2(theta) + 1# is a perfect square, if you can remember the formula #(x+y)^2 = x^2 + 2xy + y^2#

So we can say

#tan^4(theta) + 2tan^2(theta) + 1 = (tan^2(theta) + 1)^2#

However, as we know, from the pythagorean identity

#(sin^2(theta) + cos^2(theta))/cos^2(theta) = 1/cos^2(theta)#

#tan^2(theta) + 1 = sec^2(theta)#

So,

#tan^4(theta) + 2tan^2(theta) + 1 = (sec^2(theta))^2#

#tan^4(theta) + 2tan^2(theta) + 1 = sec^4(theta)#