Question

Solve the equation: \[\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{\tan 8x}}{{\tan 2x}}\]

Answer 1

Hint: To solve such questions, we take either the left hand side of the equation or right hand side of the equation and simplify it till we reach the term written in the other side. Here we take the LHS side of the equation and try to convert it into sin and cos functions, so that we can convert it into tan function easily. We convert it into tan function because RHS is tan function.
Formula used: We have used the following formulas in the solution to prove the required,
\[\tan x = \dfrac{{\sin x}}{{\cos x}}\] this is used to convert tan into sin, cos and vice versa.
\[\sin 2x = 2\sin x\cos x\]
\[\left( {1 - \cos ax} \right) = 2{\sin ^2}\dfrac{a}{2}x\].
\[\sec x = \dfrac{1}{{\cos x}}\].

Complete step-by-step solution:
Here in this question we have to prove that \[\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{\tan 8x}}{{\tan 2x}}\]. To do his first of all we will take the LHS side of the equation. So,
LHS= \[\dfrac{{sec8x - 1}}{{sec4x - 1}}\]
Since we know that \[\sec x = \dfrac{1}{{\cos x}}\], so,
\[
   \Rightarrow \dfrac{{\left( {\dfrac{1}{{\cos 8x}} - 1} \right)}}{{\left( {\dfrac{1}{{\cos 4x}} - 1} \right)}} \\
   \Rightarrow \dfrac{{\left( {1 - \cos 8x} \right)\left( {\cos 4x} \right)}}{{\left( {1 - cos4x} \right)\left( {\cos 8x} \right)}} \\
 \]
Now, here we see that in the above step we can apply the formula \[\left( {1 - \cos ax} \right) = 2{\sin ^2}\dfrac{a}{2}x\].
Using this we move ahead as,
\[ \Rightarrow \dfrac{{2\sin 4x\cos 4x\left( {\sin 4x} \right)}}{{2{{\sin }^2}2x\left( {\cos 8x} \right)}}\]
We know that \[\sin 2x = 2\sin x\cos x\], we can use this formula to reach,
\[
   \Rightarrow \dfrac{{2\sin 4x\cos 4x\left( {2\sin 2x\cos 2x} \right)}}{{2{{\sin }^2}2x\left( {\cos 8x} \right)}} \\
   \Rightarrow \dfrac{{\sin 8x\cos 2x}}{{\sin 2x\left( {\cos 8x} \right)}} \\
 \]
we know that \[\tan x = \dfrac{{\sin x}}{{\cos x}}\], we use this formula in the previous step to reach next
\[ \Rightarrow \dfrac{{\tan 8x}}{{\tan 2x}} = RHS\]
 Thus we have reached RHS from LHS. Hence we have proved the required \[\dfrac{{\sec 8x - 1}}{{\sec 4x - 1}} = \dfrac{{\tan 8x}}{{\tan 2x}}\].

Note: This is to note that this is not the only way to prove the equation. You can use any other path also to reach the destination. Before deciding to choose the way to prove results, one must understand the requirements of the part to be proven. Most common way to start your proof is by breaking the side you are starting from in sin and cos terms