Question

PT sec 8 theta-1/sec 4 theta-1=tan 8 theta/tan 2 theta

Answer 1

LHS
(sec8θ-1)/(sec4θ-1)
=(1/cos8θ-1)/(1/cos4θ-1)
=[(1-cos8θ)/cos8θ]/[(1-cos4θ)/cos4θ]
=(2sin²4θ/cos8θ)/(2sin²2θ/cos4θ)      [∵, 1-cos2θ=2sin²θ]
=2(2sin2θcos2θ)²/cos8θ×cos4θ/2sin²2θ
=4cos²2θcos4θ/cos8θ
RHS
tan8θ/tan2θ
=(sin8θ/cos8θ)/(sin2θ/cos2θ)
=2sin4θcos4θ/cos8θ×cos2θ/sin2θ
=2.2sin2θcos2θcos4θ/cos8θ×cos2θ/sin2θ
=4cos²2θcos4θ/cos8θ
∴, LHS=RHS(Proved)

Answer 2

L.H.S =sec(8x)−1sec(4x)−1

=1cos(8x)−11cos(4x)−1

=1−cos(8x)1−cos(4x)×cos(4x)cos(8x)

Since (1-cos2x=2sin^2 (x))

=2sin2(4x)2sin2(2x)×cos(4x)cos(8x)

=2sin(4x)cos(4x)cos(8x)×sin(4x)2sin2(2x)

=sin(8x)cos(8x)×2sin(2x)cos(2x)2sin2(2x)

=sin(8x)cos(8x)×cos(2x)sin(2x)

=tan(8x)×cot(2x)

=tan(8x)/tan(2x)= R.H.S