Respuesta :

AL2006
This kind of exercise can be real drudgery.  But it's almost all simple arithmetic, so better you than me.  I'll do one of these for you, which will show you how to do the other one.

a).  sin · cot / sec

You're supposed to know that [ cotangent = cosine/sine ]
and [ secant = 1/cosine ].

Then the problem becomes

         sin · (cos/sin) / (1/cos)  =  cos²

Aw shucks, I might as well also set up 'b)' for you:

b).  cos · csc / tan

You're supposed to know that [ cosecant = 1/sine ]
and  [ tangent = sine/cosine ].

Then the problem becomes

             cos · (1/sin) / (sin/cos)  =  (cos/sin)²

Now simply plug in the given values of cos A and sin A .


And may I compliment you on your nail care !