Question

hey!!
Can anybody solve this??
Sec (4A-1) ÷ Sec (2A-1)= tan4A. cotA

Answer 1

ANSWER:-
sec4A -1 / sec2A-1 

  = 1/ cos4A - 1 / 1/cos2A -1

  =  (1- cos4A ) . cos2A /  cos4A (1- cos2A )

  =  2 sin2 2A . cos2A / cos4A . 2 sin2 A

  =  2 sin2A cos2A . sin2A / cos4A . 2sin2 A

  = sin4A . 2 sinA cosA / cos4A . 2 sin2 A

  =  tan4A / tanA

=tan4A. cotA
PLEASE MARK AS BRAINLIEST

Answer 2

Answer:

Step-by-step explanation:

1. LHS= (sec4A-1)/(sec2A-1)

= {(1/cos4A)-1}/{(1/cos2A)-1}By reciprocity and solving:-                                                                                                                                              = {(1-cos4A).cos2A}/{(1-cos2A).cos4A}

= {2.sin^2 2A. cos2A}/{2Sin^2A.cos4A}

= sin2A.sin4A/ 2sin^2A.cos4A      (as 2sin2A.cos2A=sin4A)

= (2sinA.cosA.tan4A)/2sin^2A

= (cosA.tan4A)/sinA

= cotA.tan4A

= tan4A. cotA =RHS, proved║