Question

Prove \[\dfrac{\sec 8A-1}{\sec 4A-1}=\dfrac{\tan 8A}{\tan 2A}\].

Answer 1

Hint: We use
\[\dfrac{\sec 8A-1}{\sec 4A-1}=\dfrac{\tan 8A}{\tan 2A}\] …….$(1)$
$\sec A=\dfrac{1}{\cos A}$ …….$(2)$
Then as the right side involves tan we need to rewrite the left side in terms of sin and cos to get a tan. For this we will have to use the double angle formula
$1-c\cos 2A=2{{\sin }^{2}}A$ …….$(3)$

Complete step by step solution:
This is a question of trigonometry. Trigonometry is the branch of mathematics which deals with triangles, it’s sides and area. It gives relationship between sides an
We first write sec in terms of cos and then obtain after rearranging the denominators that
$\dfrac{\dfrac{1}{\cos 8A}-1}{\dfrac{1}{\cos 4A}-1}$ ……$(4)$
= $(\dfrac{1-\cos 8A}{1-\cos 4A})\dfrac{\cos 4A}{\cos 8A}$ ……$(5)$
Then we use the double angle formula from the hint to get
$(\dfrac{2{{\sin }^{2}}4A}{2{{\sin }^{2}}2A})\dfrac{\cos 4A}{\cos 8A}$ …….($6$)
We then use the double angle formula for sin i.e.
$\sin 2A=2\sin A\cos A$…….($7$)
to break one of the terms in the numerator. This gives
$(\dfrac{2\sin 2A\cos 2A\sin 4A}{{{\sin }^{2}}2A})\dfrac{\cos 4A}{\cos 8A}$ ……($8$)
$=(\dfrac{2\cos 2A\sin 4A}{\sin 2A})\dfrac{\cos 4A}{\cos 8A}$ ……($9$)
We end the calculation by collecting the 4A terms and reusing the double angle formula and then merging sin and cos as tan to give
$(\dfrac{2\sin 4A\cos 4A}{\cos 8A}\dfrac{\cos 2A}{\sin 2A})$ ……($10$)
$=(\dfrac{\sin 8A}{\cos 8A}\dfrac{\cos 2A}{\sin 2A})$ …….($11$)
$=\dfrac{\tan 8A}{\tan 2A}$ ……($12$)

Hence proved.

Note:
We could’ve also used the double angle formula to break cos as cos squared but then arriving at the result wouldn’t be so direct. Use those formula which get you close to your answer.