How to do these two problems about trigonometric identities?

1) 

a. Remember  that  the cosine is the reciprocal of the secant

So  

A / D   =  sec (theta)

1 / [ A / D]  =  1/ cos (theta)           [ flip both fractions over ]

D /A  =  cos(theta)

D / A   =  cos (theta)

b.   cos (theta)  =  1 / sec(theta)

c.  tan (theta) =  y / x  =  3 / 4     so x = 4

If sin (theta)  = y/r  =   3/5  then  r  = 5

So....cos (theta)  =  x / r   =   4/5

So we have :

cos (theta)  = D /A

4/5  =  .25 cm / A      rearrange as

A   =  .25 cm. / (4/5)   =  (1/4) / (4/5)  =  (1/4) * (5/4)  =  5/16  cm

2.

a.  sin(theta) sec(theta)  =   sin(theta) *  [1/cos(theta) ]   =  sin (theta) / cos (theta)  =

tan (theta)

b. 

The cotangent  of the angle would be   adj / opp   =  d / h

So we have

cot (theta)  =  d / h

cot  (60°)  ≈  .58

.58   =  3008 / h       rearrange as

h  =  3008 / .58   ≈   5186 ft