Question

tan4x÷tanx=sec4x-1÷sec2x+1​

Answer 1

we have to prove that tan4x/tanx = (sec4x - 1)/(sec2x - 1)

proof : RHS = (sec4x - 1)/(sec2x - 1)

= (1/cos4x - 1)/(1/cos2x - 1)

= (1 - cos4x)/(1 - cos2x) × cos2x/cos4x

using formula,

1 - cos2θ = 2sin²θ

so, 1 - cos4x = 2sin²(2x)

1 - cos2x = 2sin²x

= (2sin²2x)/(2sin²x) × (cos2x)/(cos4x)

= [2sin2x cos2x]/[cos4x] × [sin2x]/[2sin²x]

we know, 2sinθcosθ = sin2θ

so, 2sin2x cos2x = sin4x

= sin4x/cos4x × (2sinx cosx)/(2sin²x)

= tan4x × (cosx/sinx)

= tan4x × cotx

= tan4x × 1/tanx

= tan4x/tanx = RHS

Therefore, tan4x/tanx = (sec4x - 1)/(sec2x - 1)

Answer 2

Step-by-step explanation:

Here you go ! answer in attachment for sec4x-1/sec2x-1 = tan4x/tanx