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Prove that\frac{\sec 8 \alpha-1}{\sec 4 \alpha-1}=\frac{\tan 8 \alpha}{\tan 2 \alpha} |
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Answer» 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/cos8θRHStan8θ/tan2θ=(sin8θ/cos8θ)/(sin2θ/cos2θ)=2sin4θcos4θ/cos8θ×cos2θ/sin2θ=2.2sin2θcos2θcos4θ/cos8θ×cos2θ/sin2θ=4cos²2θcos4θ/cos8θ∴, LHS=RHS(Proved) I took theta on the place of alpha don't worry about that |
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