Question

# Prove that $\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{\tan 8x}}{{\tan 2x}}$

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.

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.
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.