If sum of the series $\frac {\tan 1}{\cos 2}+\frac{\tan 2}{\cos 4} +\frac{\tan 3}{\cos 6}...\frac{\tan 1024}{\cos 2048}=\tan \lambda -\tan \mu$

Question

Find the value of $\lambda +\mu$

The expression is $$\frac{\sin 1}{\cos 1 \cos 2}+\frac{\sin 2}{\cos 2 \cos 4}......$$ $$\frac {2\sin^21}{\sin 2\cos 2} +\frac{2\sin^22}{\sin 4\cos 4}$$ and so on, but that isn’t getting me anywhere. What can be done?

I am asking where you got the question. (See the guidelines for asking good questions here.) — GEdgar, Dec 21 '19 at 12:21
My teacher gave this to all the students. I don’t know where it comes from — Aditya, Dec 21 '19 at 12:33
You can check the method in Bellow link https://www.mathsdiscussion.com/forum/topic/method-of-difference/?part=1#postid-53 For your question n=1024 — mathsdiscussion.com, Dec 21 '19 at 12:20

lab bhattacharjee · Accepted Answer · 2019-12-21T12:42:38.540

5

If $\tan x,\tan2x$ are non-zero and finite, i.e., we need $2x\ne n90^\circ$ where $n$ is any integer

$$\dfrac{\tan x}{\cos2x}=\dfrac{\sin x}{\cos x\cos2x}=\dfrac{\sin(2x-x)}{\cos x\cos2x}=\tan2x-\tan x$$

So the problem should be $$\sum_{r=0}^{10}\dfrac{\tan(2^rx)}{\cos(2^{r+1}x)}$$

edited Dec 21 '19 at 12:42

answered Dec 21 '19 at 12:31

lab bhattacharjee

274,582

score 3 · Answer 2 · answered Dec 21 '19 at 12:27

3

Hint:$\frac{1}{\cos(2x)} = \frac{\tan(2x)}{\tan(x)}-1$

Try proving this yourself. After this it is just simple telescoping series

answered Dec 21 '19 at 12:27

user600016

2,165

1

This is the same as Alok's solution referenced by @mathdiscusscon.com, but it doesn't telescope. $(\tan 2 - \tan 1) + (\tan 4 - \tan 2) + (\tan 6 - \tan 3) +\dots$ – GEdgar Dec 21 '19 at 12:43

If sum of the series $\frac {\tan 1}{\cos 2}+\frac{\tan 2}{\cos 4} +\frac{\tan 3}{\cos 6}...\frac{\tan 1024}{\cos 2048}=\tan \lambda -\tan \mu$

2 Answers2