\((\sec 8A – 1)/(\sec 4A – 1)\) is Equal to What?
Question
\[ \frac{\sec 8A-1}{\sec 4A-1} \] is equal to
(a) \(\dfrac{\tan 2A}{\tan 8A}\)
(b) \(\dfrac{\tan 8A}{\tan 2A}\)
(c) \(\dfrac{\cot 8A}{\cot 2A}\)
(d) none of these
Solution
Use the identity:
\[ \sec\theta-1 =\frac{1-\cos\theta}{\cos\theta} \]
Therefore,
\[ \frac{\sec 8A-1}{\sec 4A-1} = \frac{\frac{1-\cos 8A}{\cos 8A}} {\frac{1-\cos 4A}{\cos 4A}} \]
\[ = \frac{1-\cos 8A}{1-\cos 4A} \cdot \frac{\cos 4A}{\cos 8A} \]
Using \(1-\cos\theta=2\sin^2(\theta/2)\),
\[ = \frac{2\sin^2 4A}{2\sin^2 2A} \cdot \frac{\cos 4A}{\cos 8A} \]
\[ = \frac{\sin^2 4A}{\sin^2 2A} \cdot \frac{\cos 4A}{\cos 8A} \]
Since \[ \sin 4A=2\sin 2A\cos 2A \]
\[ = \frac{(2\sin 2A\cos 2A)^2}{\sin^2 2A} \cdot \frac{\cos 4A}{\cos 8A} \]
\[ = 4\cos^2 2A \cdot \frac{\cos 4A}{\cos 8A} \]
Using \[ 2\cos^2 2A=1+\cos 4A \]
\[ 4\cos^2 2A=2(1+\cos 4A) \]
\[ = \frac{2(1+\cos 4A)\cos 4A}{\cos 8A} \]
\[ = \frac{4\cos^2 4A\cos 2A}{\cos 8A} = \frac{\tan 8A}{\tan 2A} \]
Hence,
\[ \frac{\sec 8A-1}{\sec 4A-1} = \frac{\tan 8A}{\tan 2A} \]
Final Answer
\[ \boxed{\frac{\tan 8A}{\tan 2A}} \]
Therefore, the correct option is (b).