Math, asked by leenaj499, 8 months ago

Pl solve l.h.s.
Sec8theta-1/sec4theta-1= tan8theta/tan2theta

Pl solve fast

Answers

Answered by Anonymous
18

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)

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