Question

The value of \[\dfrac{\sec 8A-1}{\sec 4A-1}\] is equal to
(a) \[\dfrac{\tan 2A}{\tan 8A}\]
(b) \[\dfrac{\tan 8A}{\tan 2A}\]
(c) \[\dfrac{\cot 8A}{\cot 2A}\]
(d) \[\dfrac{\tan 6A}{\tan 2A}\]

Answer 1

Hint: We solve this problem by using the simple formulas of trigonometric ratios and half angle formulas. We have the relation between secant and cosine trigonometric ratios as
\[\Rightarrow \sec \theta =\dfrac{1}{\cos \theta }\]
Now, we have the formulas of half angles for cosine and sine ratios as
\[\Rightarrow \cos 2\theta =1-2{{\sin }^{2}}\theta \]
\[\Rightarrow \sin 2\theta =2\sin \theta \cos \theta \]
We also have the formula for tangent trigonometric ratio as
\[\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }\]
By using the above formulas we find the required value.

Complete step by step answer:
We are asked to find the value of \[\dfrac{\sec 8A-1}{\sec 4A-1}\]
Let us assume that the required value of given trigonometric function as
\[\Rightarrow x=\dfrac{\sec 8A-1}{\sec 4A-1}\]
We know that the relation between secant and cosine trigonometric ratios as
\[\Rightarrow \sec \theta =\dfrac{1}{\cos \theta }\]
By using this relation to above equation we get
\[\begin{align}
  & \Rightarrow x=\dfrac{\dfrac{1}{\cos 8A}-1}{\dfrac{1}{\cos 4A}-1} \\
 & \Rightarrow x=\left( \dfrac{1-\cos 8A}{1-\cos 4A} \right)\left( \dfrac{\cos 4A}{\cos 8A} \right) \\
\end{align}\]
We know that the half angle formula for cosine ratio that is
\[\begin{align}
  & \Rightarrow \cos 2\theta =1-2{{\sin }^{2}}\theta \\
 & \Rightarrow 1-\cos 2\theta =2{{\sin }^{2}}\theta \\
\end{align}\]
Now, by using this half angle formula for the above equation for the first term we get
\[\Rightarrow x=\left( \dfrac{2{{\sin }^{2}}4A}{2{{\sin }^{2}}2A} \right)\left( \dfrac{\cos 4A}{\cos 8A} \right)\]
Now, let us rearrange the terms in the above equation such that we can get the half angle formula for sine ratio then we get
\[\Rightarrow x=\left( \dfrac{2\sin 4A.\cos 4A}{\cos 8A} \right)\left( \dfrac{\sin 4A}{2{{\sin }^{2}}2A} \right)\]
We know that the half angle formula for the sine ratio that is
\[\Rightarrow \sin 2\theta =2\sin \theta \cos \theta \]
By using this formula of sine ratio to above equation we get
\[\begin{align}
  & \Rightarrow x=\left( \dfrac{\sin 8A}{\cos 8A} \right)\left( \dfrac{2\sin 2A.\cos 2A}{2{{\sin }^{2}}2A} \right) \\
 & \Rightarrow x=\left( \dfrac{\sin 8A}{\cos 8A} \right)\left( \dfrac{\cos 2A}{\sin 2A} \right) \\
\end{align}\]

We know that the formula for the tangent trigonometric ratio as
\[\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }\]
By using this formula to above equation then we get
\[\begin{align}
  & \Rightarrow x=\left( \dfrac{\sin 8A}{\cos 8A} \right)\left( \dfrac{1}{\left( \dfrac{\sin 2A}{\cos 2A} \right)} \right) \\
 & \Rightarrow x=\dfrac{\tan 8A}{\tan 2A} \\
\end{align}\]
Therefore we can write the value of given trigonometric function as
\[\therefore \dfrac{\sec 8A-1}{\sec 4A-1}=\dfrac{\tan 8A}{\tan 2A}\]

So, the correct answer is “Option b”.

Note: Sometimes we may be asked to represent the given function in terms of cotangent trigonometric ratio.
Here we have the value of given function in terms of tangent as
\[\Rightarrow \dfrac{\sec 8A-1}{\sec 4A-1}=\dfrac{\tan 8A}{\tan 2A}\]
This value can be represented in terms of cotangent by using the relation between tangent and cotangent that is
\[\Rightarrow \tan \theta =\dfrac{1}{\cot \theta }\]
By using this formula we get
\[\begin{align}
  & \Rightarrow \dfrac{\sec 8A-1}{\sec 4A-1}=\dfrac{\left( \dfrac{1}{\cot 8A} \right)}{\left( \dfrac{1}{\cot 2A} \right)} \\
 & \Rightarrow \dfrac{\sec 8A-1}{\sec 4A-1}=\dfrac{\cot 2A}{\cot 8A} \\
\end{align}\]
This is also the correct answer.