Question

Can anyone can prove this..

Answer 1

\[\cot 4 \theta = \frac{ \cot^4\theta - 6\cot ^2 +1 }{ 4\cot \theta(\cot^2\theta-1)}\]

Answer 2

@thadyoung

Answer 3

are you allowed to use DeMovire's Theorem ?

Then one of the method will be to find cos 4thetha and sin 4thetha, then divide...

Answer 4

Using
\(cot(2x) = \frac{cot^2(x)-1}{2cot(x)}\)
and expand twice gives the right hand side.

Answer 5

Using \(cot(2x) = \frac{cot^2(x)-1}{2cot(x)}\)
\[\large cot(4x) = \frac{(\frac{cot^2(x)-1}{2cot(x)})^2 -1}{2(\frac{cot^2(x)-1}{2cot(x)})}\]
\[\Large =\frac{\frac{( cot^2(x)-1)^2-4cot^2(x)}{4cot^2(x)}} {\frac{2( cot^2(x)-1)}{2cot(x)}}\]
\[\large =\frac{cot^4(x)-6cot(x)+1}{2cot^2(x)~2(cot^2(x)-1)}\]
\[\large =\frac{cot^4(x)-6cot^2(x)+1}{4cot(x)(cot^2(x)-1)}\]