MCQ
$\frac{{\sec \,8\theta - 1}}{{\sec \,4\theta - 1}}$ is equal to
- A$tan\, 2\theta \,cot \,8\theta$
- B$tan \,8\theta\, tan \,2\theta$
- C$cot\, 8\theta \,cot \,2\theta$
- ✓$tan \,8\theta\, cot\, 2\theta$
$=\frac{2 \sin ^{2} 4 \theta}{\cos 8 \theta} \times \frac{\cos 4 \theta}{2 \sin ^{2} 2 \theta} \quad\left[\because 1-\cos 8 \theta=2 \sin ^{2} \frac{8 \theta}{2}=2 \sin ^{2} 4 \theta\right]$ and
$\left[\because 1-\cos 4 \theta=2 \sin ^{2} \frac{4 \theta}{2}=2 \sin ^{2} 2 \theta\right]$
$=\frac{(2 \sin 4 \theta \cos 4 \theta)}{\cos 8 \theta} \times \frac{\sin 4 \theta}{2 \sin ^{2} 2 \theta}$
$=\left(\frac{2 \sin 4 \theta \cos 4 \theta}{\cos 8 \theta}\right) \times\left(\frac{2 \sin 2 \theta \cos 2 \theta}{2 \sin ^{2} 2 \theta}\right)$
$=\left(\frac{\sin 2(4 \theta)}{\cos 8 \theta}\right) \times\left(\frac{\cos 2 \theta}{\sin 2 \theta}\right)=\left(\frac{\sin 8 \theta}{\cos 8 \theta}\right) \times\left(\frac{\cos 2 \theta}{\sin 2 \theta}\right)$
$\tan 8 \theta \cot 2 \theta$
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