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

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

& \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{aligned}$$ We know that the half angle formula for cosine ratio that is $$\begin{aligned} & \Rightarrow \cos 2\theta =1-2{{\sin }^{2}}\theta \\\ & \Rightarrow 1-\cos 2\theta =2{{\sin }^{2}}\theta \\\ \end{aligned}$$ Now, by using this half angle formula for the above equation for the first term we get $$\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{aligned} & \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{aligned}$$ We know that the formula for the tangent trigonometric ratio as $$\Rightarrow \tan \theta =\dfrac{\sin \theta }{\cos \theta }$$ By using this formula to above equation then we get $$\begin{aligned} & \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{aligned}$$ Therefore we can write the value of given trigonometric function as $$\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{aligned} & \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{aligned}$$ This is also the correct answer.