Question

Prove that (sec 12A-1)/(sec 6A-1)=tan 12A/tan 3A

Answer 1

Let $x=3A$. Recall the following identities,

$\cos^2\theta=\dfrac{1+\cos2\theta}2$

$\sin^2\theta=\dfrac{1-\cos2\theta}2$

$\sin2\theta=2\sin\theta\cos\theta$

Now,

$\dfrac{\sec12A-1}{\sec6A-1}=\dfrac{\sec4x-1}{\sec2x-1}$

$=\dfrac{\cos2x(1-\cos4x)}{\cos4x(1-\cos2x)}$

$=\dfrac{2\cos2x\sin^22x}{\cos4x(1-\cos2x)}$

$=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x(1-\cos^22x)}=\dfrac{2\cos2x\sin^22x(1+\cos2x)}{\cos4x\sin^22x}=\dfrac{2\cos2x(1+\cos2x)}{\cos4x}$

$=\dfrac{4\cos2x\cos^2x}{\cos4x}$

$=\dfrac{4\cos2x\cos^2x\sin4x}{\cos4x\sin4x}=\dfrac{4\cos2x\cos^2x\tan4x}{\sin4x}$

$=\dfrac{4\cos2x\cos^2x\tan4x}{2\sin2x\cos2x}=\dfrac{2\cos^2x\tan4x}{\sin2x}$

$=\dfrac{2\cos^2x\tan4x}{2\sin x\cos x}=\dfrac{\cos x\tan4x}{\sin x}$

$=\dfrac{\tan4x}{\frac{\sin x}{\cos x}}=\dfrac{\tan4x}{\tan x}=\dfrac{\tan12A}{\tan3A}$

QED