Question

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

Answer 1

How about using these first:

\[cosec(4A) = \frac{1}{\sin(4A)}\]
\[\cot(4A) = \frac{\cos(4A)}{\sin(4A)}\]

Answer 2

So LHS becomes :

\[LHS \implies \frac{1 + \cos(4A)}{\sin(4A)}\]

Answer 3

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

Answer 4

Got it..

Answer 5

You want to just remember these formulas here:

\[\tan^2(u) = \frac{1 - \cos(2u)}{1 + \cos(2u)}\]
\[\tan(2u) = \frac{2\tan(u)}{1 - \tan^2(u)}\]

Answer 6

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...

Answer 7

By using identity \(sin^2(x) + cos^2(x) = 1\) :

Sin(4A) can be written as:

\[\sin(4A) = \sqrt{1 - \cos^2(4A)} \implies \sqrt{1 - \cos(4A)} \cdot \sqrt{1 +\cos(4A)}\]

Answer 8

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

Answer 9

I too... :P

Answer 10

So, moving forward:

\[\frac{1 + \cos(4A)}{\sqrt{1 + \cos(4A)} \sqrt{1 - \cos(4A)}} \implies \frac{\sqrt{1 + \cos(4A)}}{{\sqrt{1 - \cos(4A)}}}\]

Answer 11

Now just take reciprocal, you will get the same form:

\[\huge \frac{1}{\frac{\sqrt{1 - \cos(4A)}}{\sqrt{1 + \cos(4A)}}}\]

Answer 12

Now use the 1 formula:

\[\implies \frac{1}{\sqrt{\tan^2(2A)}} \implies \frac{1}{\tan(2A)}\]

Answer 13

Use the second formula here and you will get your answer...

Answer 14

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

Answer 15

yea :P

Answer 16

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.. :(

Answer 17

yes

Answer 18

i know them :)

Answer 19

Just divide them.. :)

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

ahh yes i can :) thank you :)

Answer 24

You are welcome dear..