Prove the following identity: $\dfrac{sin4A}{cos2A} × \dfrac{1 - cos2A}{1 - cos4A} = tanA$

Solution:

LHS

$= \dfrac{sin4A}{cos2A} × \dfrac{1 - cos2A}{1 - cos4A}$

$= \dfrac{2sin2Acos2A}{cos2A} × \dfrac{1 - cos2A}{1 - cos4A}$

$= 2sin2A × \dfrac{1 - cos2A}{1 - cos4A}$

Remember this identity: (1 - cos2A) = 2sin²A

$= 2sin2A × \dfrac{2sin²A}{1 - cos(2×2A)}$

Again use the same above mentioned identity

$= 2sin2A × \dfrac{2sin²A}{2sin²2A}$

$= \dfrac{2sin2A}{2sin²2A} × 2sin²A$

$= \dfrac{1}{sin2A} × 2sin²A$

Use the multiple angles formula of sin2A = 2sinAcosA

$= \dfrac{1}{2sinAcosA} × 2sin²A$

$= \dfrac{sinA}{cosA}$

$= tanA$

RHS

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