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\frac { \sec 8 \theta - 1 } { \ sin 4\theta - 1 } = \frac { \tan 8 \theta } { \tan 20 } |
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Answer» (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/cos8θRHStan8θ/tan2θ=(sin8θ/cos8θ)/(sin2θ/cos2θ)=2sin4θcos4θ/cos8θ×cos2θ/sin2θ=2.2sin2θcos2θcos4θ/cos8θ×cos2θ/sin2θ=4cos²2θcos4θ/cos8θ∴, LHS=RHS(Proved) |
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