Question

Prove that

(tan4A + tan2A)(1-tan²3Atan²A) = 2tan3Asec²A

Answer 1

tan(3A+A) = (tan3A + tanA)/(1-tan3AtanA)

tan(3A-A) = (tan3A - tanA)/(1+tan3AtanA)

add them, and you have a common
denominator of (1-tan²3Atan²A)
numerator = (tan3A + tanA)(1+tan3AtanA) + (tan3A - tanA)(1-tan3AtanA)
= 2(tan3A + tan3Atan²A)
= 2tan3A(1+tan²A)
= 2tan3Asec²2A

Answer 2

tan(3A+A) = (tan3A + tanA)/(1-tan3AtanA)

tan(3A-A) = (tan3A - tanA)/(1+tan3AtanA)

add them, and you have a common
denominator of (1-tan²3Atan²A)
numerator = (tan3A + tanA)(1+tan3AtanA) + (tan3A - tanA)(1-tan3AtanA)
= 2(tan3A + tan3Atan²A)
= 2tan3A(1+tan²A)
= 2tan3Asec²2A