Dear student sec 8θ-1sec 4θ-1Consider the numerator, sec 8θ-1=1cos 8θ-1=1-cos 8θcos 8θ=2sin24θcos8θ=2sin24θcos 8θ×sin 8θsin 8θ=2sin24θsin 8θ×tan 8θ=2sin24θ2sin 4θ cos 4θ×tan 8θ=sin 4θcos 4θ×tan 8θ=tan 4θ tan 8θSimilarly, we get sec 4θ-1 = tan 4θ tan 2θNow, sec 8θ-1sec 4θ-1=tan 4θ tan 8θtan 2θ tan 4θ=tan 8θtan 2θ Regards

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\frac{\sec 8 θ - 1}{\sec 4 θ - 1} Consider the numerator, \sec 8 θ - 1 = \frac{1}{\cos 8 θ} - 1 = \frac{1 - \cos 8 θ}{\cos 8 θ} = \frac{2 \sin^{2} 4 θ}{\cos 8 θ} = \frac{2 \sin^{2} 4 θ}{\cos 8 θ} \times \frac{\sin 8 θ}{\sin 8 θ} = \frac{2 \sin^{2} 4 θ}{\sin 8 θ} \times \tan 8 θ = \frac{2 \sin^{2} 4 θ}{2 \sin 4 θ \cos 4 θ} \times \tan 8 θ = \frac{\sin 4 θ}{\cos 4 θ} \times \tan 8 θ = \tan 4 θ . \tan 8 θ Similarly, we get \sec 4 θ - 1 = \tan 4 θ . \tan 2 θ Now, \frac{\sec 8 θ - 1}{\sec 4 θ - 1} = \frac{\tan 4 θ . \tan 8 θ}{\tan 2 θ . \tan 4 θ} = \frac{\tan 8 θ}{\tan 2 θ}

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