Question

csc h/ (tan h + cot h). Express in terms of sin h and cos h and simplify.

Answer 1

its just normal trig

Answer 2

hyperbolics behave the exact same as normal trig functions

Answer 3

csc h = 1/sin h
tan h = sin h /cos h
cot h = cos h/ sin h

= (1/sin h)/(sin/cos+cos/sin)
=(1/sin)/(sin^2 + cos^2 / sin cos)
=(1/sin)/(1/sin cos)
=cos h//