Question

Prove that \[\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{\tan 8x}}{{\tan 2x}}\]

Answer 1

Hint: In the above given question we can proceed using the concepts that, \[\cos x = \dfrac{1}{{\sec x}}\]and also \[\cos 2x = 1 - 2{\sin ^2}x\]and substituting all this in the above values we can prove the given.

Complete step-by-step answer:
As the given is \[\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}}\], using the formula of\[\cos x = \dfrac{1}{{\sec x}}\] and substituting it in the previous equation, we get,
\[ \Rightarrow \]\[\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{\dfrac{1}{{\cos 8x}} - 1}}{{\dfrac{1}{{\cos 4x}} - 1}}\]
On simplifying further we get,
\[ \Rightarrow \]\[\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{\dfrac{{(1 - \cos 8x)}}{{\cos 8x}}}}{{\dfrac{{(1 - \cos 4x)}}{{\cos 4x}}}}\]
\[ \Rightarrow \dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{(1 - \cos 8x)\cos 4x}}{{\cos 8x(1 - \cos 4x)}}\]
Now, simplify with the formula of \[{\text{cos2x = 1 - 2si}}{{\text{n}}^{\text{2}}}{\text{x}}\]
So,
\[
  \cos 4x = 1 - 2{\sin ^2}2x \\
  \cos 8x = 1 - 2{\sin ^2}4x \\
  \]
Now substituting the above,
\[ = \dfrac{{(1 - (1 - 2{{\sin }^2}4x))\cos 4x}}{{\cos 8x(1 - (1 - 2{{\sin }^2}2x))}}\]
On simplification we get,
\[ = \dfrac{{(2{{\sin }^2}4x)\cos 4x}}{{\cos 8x(2{{\sin }^2}2x)}}\]
Now on splitting we get,
\[ = \dfrac{{(2\sin 4x)(\sin 4x)\cos 4x}}{{\cos 8x(2\sin 2x)(\sin 2x)}}\]
Now, use the formula of \[{\text{sin2x = 2sinxcosx}}\] in the above equation,
\[ = \dfrac{{(2\sin 4x)(2\sin 2x\cos 2x)\cos 4x}}{{\cos 8x(2\sin 2x)(\sin 2x)}}\]
On cancelling common terms we get,
\[ = \dfrac{{(2\sin 4x)(\cos 2x)\cos 4x}}{{\cos 8x(\sin 2x)}}\]
Rearranging the terms of above equation, we get,
\[ = \dfrac{{(2\sin 4x\cos 4x)}}{{\cos 8x(\dfrac{{\sin 2x}}{{\cos 2x}})}}\]
Now on using \[{\text{sin2x = 2sinxcosx}}\], we get,
\[ = \dfrac{{(\sin 8x)}}{{\cos 8x(\dfrac{{\sin 2x}}{{\cos 2x}})}}\]
Now using the trigonometry ratios \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]. The above expression can be simplified up to,
\[ = \dfrac{{\tan 8x}}{{\tan 2x}}\]
Hence, \[\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{\tan 8x}}{{\tan 2x}}\]
Hence, proved.
Additional information:
Manufacturing Industry. Trigonometry plays a major role in industry, where it allows manufacturers to create everything from automobiles to zigzag scissors. Engineers rely on trigonometric relationships to determine the sizes and angles of mechanical parts used in machinery, tools and equipment.

Note: Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. Use trigonometric conversion formulas to reach the final step.\[\cos x = \dfrac{1}{{\sec x}}\] and \[\cos 2x = 1 - 2{\sin ^2}x\] are required to solve the given problem.