Question

Establish the identity, proving the lhs equals the rhs without making any changes to the right

Will fan and medal

Answer 1

\[\cot(2\theta)=\cot^2\theta-1/2\cot \theta\]

Answer 2

Hmm I'd probably go with sines and cosines on this one.\[\Large\rm \cot(2\theta)=\frac{\cos(2\theta)}{\sin(2\theta)}\]And then apply our Double Angle Formulas.

Answer 3

\[\Large\rm =\frac{\cos^2\theta-\sin^2\theta}{2\sin \theta \cos \theta}\]

Answer 4

From there it requires a bit of a tricky step I guess...
Divide top and bottom by sin^2(theta)

Answer 5

\[\Large\rm \frac{\cos^2\theta-\sin^2\theta}{2\sin \theta \cos \theta}\left(\frac{\frac{1}{\sin^2\theta}}{\frac{1}{\sin^2\theta}}\right)=\frac{\frac{\cos^2\theta}{\sin^2\theta}-\frac{\sin^2\theta}{\sin^2\theta}}{2\cancel{\sin \theta}\cos \theta\cdot \frac{1}{\sin^{\cancel 2}\theta}}\]

Answer 6

Understand why I did that? :o
Aw you went offline :c sads

Answer 7

pity you missed the brackets....(cot^2 theta-1)