#tan theta + 2tan2theta + 4cot4theta = cot theta # How to prove?
1 Answer
We seek to prove that:
# tan theta + 2tan2theta + 4cot4theta -= cot theta #
Let us start with the LHS, as this is the most complex part of the expression:
# LHS = tan theta + 2tan2theta + 4cot4theta #
# \ \ \ \ \ \ \ \ = tan theta + 2tan2theta + 4/(tan4theta) #
We can use the tangent double angle formula:
# tan 2A -= (2tanA)/(1-tan^2A) #
So then:
# LHS = tan theta + 2tan2theta + 4/(tan2(2theta)) #
# \ \ \ \ \ \ \ \ = tan theta + 2tan2theta + 4(1-tan^2 2theta)/(2tan 2theta) #
# \ \ \ \ \ \ \ \ = tan theta + 2tan2theta + 2(1-tan^2 2theta)/(tan 2theta) #
# \ \ \ \ \ \ \ \ = (tan thetatan2 theta + 2tan^2 2theta + 2-2tan^2 2theta ) /(tan 2theta) #
# \ \ \ \ \ \ \ \ = (tan thetatan2 theta + 2) /(tan 2theta) #
# \ \ \ \ \ \ \ \ = ((2tan^2 theta)/(1-tan^2 theta) + 2) /((2tan theta)/(1-tan^2 theta)) #
# \ \ \ \ \ \ \ \ = ( (2tan^2 theta + 2(1-tan^2 theta)) / (1-tan^2 theta) ) /((2tan theta)/(1-tan^2 theta)) #
# \ \ \ \ \ \ \ \ = ( (2tan^2 theta + 2-2tan^2 theta) / (1-tan^2 theta) ) /((2tan theta)/(1-tan^2 theta)) #
# \ \ \ \ \ \ \ \ = (2)/(2tan theta) #
# \ \ \ \ \ \ \ \ = 1/tan theta #
# \ \ \ \ \ \ \ \ = cot theta #
# \ \ \ \ \ \ \ \ = RHS \ \ \ \ # QED