Question: Prove the following trigonometric identity:
$1 + tan 4A.tan 2A = sec 4A$


Solution:

Taking LHS

$ = 1 + tan 4A.tan 2A$

$= 1 + tan(2×2A) . tan2A$

Remember: $tan2A = \dfrac{2 tanA}{1 - tan^2 A}$

$= 1 + \dfrac{2tan2A}{1 - tan^2 2A} . tan2A$

$= 1 + \dfrac{2 tan^2 2A}{1 - tan^2 2A}$

$= \dfrac{(1 - tan^2 2A) + 2tan^2 2A}{1 - tan^2 2A}$

$= \dfrac{1 + tan^2 2A}{1 - tan^2 2A}$

Remember: $cos 2A = \dfrac{1 - tan^2A}{1 +tan^2 A}$

$= \dfrac{1}{cos(2×2A)}$

$= \dfrac{1}{cos 4A}$

$= sec 4A$

= RHS

#proved

Other Solutions from Multiple Angles

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