Question

Sec 8a_1 ÷ sec 4a _ 1 = tan 8a÷ tan2a

Answer 1

we have to prove that, (sec8a - 1)/(sec4a - 1) = tan8a/tan2a

proof : LHS = (sec8a - 1)/(sec4a - 1)

= (1/cos8a - 1)/(1/cos4a - 1)

= (1 - cos8a)/(1 - cos4a) × cos4a/cos8a

using formula, 1 - cos2θ = 2sin²θ

= (2sin²4a)/(2sin²2a) × cos4a/cos8a

= (2sin4a cos4a)/cos8a × (sin4a)/(2sin²2a)

using formula, 2sinx cosx = sin2x

= sin8a/cos8a × (2sin2a cos2a)/(2sin²2a)

= tan8a × (cos2a)/(sin2a)

= tan8a × cot2a

= tan8a/tan2a = RHS

hence proved