## AonZ Group Title Prove cosec4A + cot4A = 1/2 (cotA- tanA) Any way of doing this easily? like with minimum working out? @jim_thompson5910 cause i get stuck with a massive page of working out one year ago one year ago

1. waterineyes

How about using these first: $cosec(4A) = \frac{1}{\sin(4A)}$ $\cot(4A) = \frac{\cos(4A)}{\sin(4A)}$

2. waterineyes

So LHS becomes : $LHS \implies \frac{1 + \cos(4A)}{\sin(4A)}$

3. AonZ

yea i got that too... then i prob stuffed up somewhere...

4. waterineyes

Got it..

5. waterineyes

You want to just remember these formulas here: $\tan^2(u) = \frac{1 - \cos(2u)}{1 + \cos(2u)}$ $\tan(2u) = \frac{2\tan(u)}{1 - \tan^2(u)}$

6. waterineyes

So in the numerator above there is one 1+ cos(4A) is there but we need there one (1-cos(4A), so that we can apply the formula 1 there...

7. waterineyes

By using identity $$sin^2(x) + cos^2(x) = 1$$ : Sin(4A) can be written as: $\sin(4A) = \sqrt{1 - \cos^2(4A)} \implies \sqrt{1 - \cos(4A)} \cdot \sqrt{1 +\cos(4A)}$

8. AonZ

never seen this before :P $\huge\tan^2(u) = \frac{1 - \cos(2u)}{1 + \cos(2u)}$

9. waterineyes

I too... :P

10. waterineyes

11. waterineyes

So, moving forward: $\frac{1 + \cos(4A)}{\sqrt{1 + \cos(4A)} \sqrt{1 - \cos(4A)}} \implies \frac{\sqrt{1 + \cos(4A)}}{{\sqrt{1 - \cos(4A)}}}$

12. waterineyes

Now just take reciprocal, you will get the same form: $\huge \frac{1}{\frac{\sqrt{1 - \cos(4A)}}{\sqrt{1 + \cos(4A)}}}$

13. waterineyes

Now use the 1 formula: $\implies \frac{1}{\sqrt{\tan^2(2A)}} \implies \frac{1}{\tan(2A)}$

14. waterineyes

15. waterineyes

Oh, I forget, you need verification of that formula???

16. AonZ

yea :P

17. waterineyes

You know this?? $\sin^2(u) = \frac{1 - \cos(2u)}{2}$ $\cos^2(u) = \frac{1 + \cos(2u)}{2}$ Now don't say you have not seen them too.. :(

18. AonZ

yes

19. AonZ

i know them :)

20. waterineyes

Just divide them.. :)

21. waterineyes

You will get formula for $$tan^2(u)$$.

22. waterineyes

$\huge \frac{\sin^2(u)}{\cos^2(u)} = \frac{\frac{1 - \cos(2u)}{2}}{\frac{1 + \cos(2u)}{2}}$ $\tan^2(u) = \frac{1- \cos(2u)}{1 + \cos(2u)}$

23. waterineyes

Now I think you can prove it.. Go ahead..

24. AonZ

ahh yes i can :) thank you :)

25. waterineyes

You are welcome dear..