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AleksAgata [21]
2 years ago
5

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

Mathematics
1 answer:
adelina 88 [10]2 years ago
5 0

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

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