Question

cos theta*cot(90-theta)=?

Answer 1

\[\cot (90-\theta) = 1/\tan(90 - \theta)\]

\[\tan (90 - \theta) = a/o\]

then \[1/\tan(90 - \theta) = o/a\]

so the problem now become

\[\cos(\theta) \times \tan(\theta)\]

and using the tan property

\[\tan = \sin/\cos\]

the problem is now

\[\cos(\theta) \times \sin(\theta)/\cos(\theta) = \sin(\theta)\]

Hope this helps