Respuesta :
Let [tex]x=3A[/tex]. Recall the following identities,
[tex]\cos^2\theta=\dfrac{1+\cos2\theta}2[/tex]
[tex]\sin^2\theta=\dfrac{1-\cos2\theta}2[/tex]
[tex]\sin2\theta=2\sin\theta\cos\theta[/tex]
Now,
[tex]\dfrac{\sec12A-1}{\sec6A-1}=\dfrac{\sec4x-1}{\sec2x-1}[/tex]
[tex]=\dfrac{\cos2x(1-\cos4x)}{\cos4x(1-\cos2x)}[/tex]
[tex]=\dfrac{2\cos2x\sin^22x}{\cos4x(1-\cos2x)}[/tex]
[tex]=\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}[/tex]
[tex]=\dfrac{4\cos2x\cos^2x}{\cos4x}[/tex]
[tex]=\dfrac{4\cos2x\cos^2x\sin4x}{\cos4x\sin4x}=\dfrac{4\cos2x\cos^2x\tan4x}{\sin4x}[/tex]
[tex]=\dfrac{4\cos2x\cos^2x\tan4x}{2\sin2x\cos2x}=\dfrac{2\cos^2x\tan4x}{\sin2x}[/tex]
[tex]=\dfrac{2\cos^2x\tan4x}{2\sin x\cos x}=\dfrac{\cos x\tan4x}{\sin x}[/tex]
[tex]=\dfrac{\tan4x}{\frac{\sin x}{\cos x}}=\dfrac{\tan4x}{\tan x}=\dfrac{\tan12A}{\tan3A}[/tex]
QED
With the help of trigonometric identities, [tex]\rm 2*cos ^2 \theta = 1 +cos\ 2 \theta[/tex], [tex]\rm 2*sin^2 \theta = 1 -cos\ 2 \theta[/tex], and [tex]\rm sin \ 2 \theta = 2\ sin\ \theta \ cos\ \theta[/tex]. We have proven that the left-hand side is equal to the right-hand side.
What is trigonometry?
Trigonometry deals with the relationship between the sides and angles of a right-angle triangle.
[tex]\rm \dfrac{sec\ 12A - 1}{sec\ 6A - 1} = \dfrac{tan\ 12A}{tan\ 3A}[/tex]
Let x = 3A
We know that the identities
[tex]\rm cos ^2 \theta = \dfrac{1 +cos\ 2 \theta}{2}[/tex]
[tex]\rm sin^2 \theta = \dfrac{1 -cos\ 2 \theta}{2}[/tex]
[tex]\rm sin \ 2 \theta = 2\ sin\ \theta \ cos\ \theta[/tex]
From the left-hand side, we have
[tex]\rm\rightarrow \dfrac{sec\ 12A - 1}{sec\ 6A - 1} \\\\\\\rightarrow \dfrac{sec\ 4x - 1}{sec\ 2x - 1} \\\\\\\rightarrow \dfrac{cos2x(1-cos 4x)}{cos 4x (1-cos2x)}\\\\\\\rightarrow \dfrac{2cos2x * sin^22x}{cos4x(1-cos2x)}[/tex]
[tex]\rm \rightarrow \dfrac{2cos 2x *sin ^22x(1+cos2x)}{cos 4x (1-cos^2x)}\\\\\\\rightarrow \dfrac{2cos 2x *sin ^22x(1+cos2x)}{cos 4x*sin^22x}\\\\\\\rightarrow \dfrac{2cos 2x (1+cos2x)}{cos 4x}\\\\\\\rightarrow \dfrac{4cos 2x* cos^2x}{cos 4x}\\\\\\[/tex]
[tex]\rm \rightarrow \dfrac{4cos2x*cos^2x* sin 4x}{cos 4x*sin4x}\\\\\\\rightarrow \dfrac{4cos2x*cos^2x* tan 4x}{sin4x}\\\\\\\rightarrow \dfrac{4cos2x*cos^2x* tan 4x}{2sin2x*cos2x}\\\\\\\rightarrow \dfrac{2cos^2x* tan 4x}{sin2x}[/tex]
[tex]\rm \rightarrow \dfrac{2cos^2x* tan 4x}{2sinx*cosx}\\\\\\\rightarrow \dfrac{cosx* tan 4x}{sinx}\\\\\\\rightarrow \dfrac{tan 4x}{\frac{sinx}{cosx}}\\\\\\\rightarrow \dfrac{tan 4x}{tanx}\\\\\\\rightarrow \dfrac{tan 12A}{tan3A} = right \ hand \ side[/tex]
Hence proved, [tex]\rm \dfrac{sec\ 12A - 1}{sec\ 6A - 1} = \dfrac{tan\ 12A}{tan\ 3A}[/tex].
More about the trigonometry link is given below.
https://brainly.com/question/22698523
