Verified answer

Hello

Using this formula:

cos(a-b) = cos[a]cos[b] + sin[a]sin[b]

cos(8)cos(38) + sin(8)sin(38 = cos (8-30)

cos (8-30) = cos (-30)

cos (-30) = cos (30) (identity)

cos (30) = [sqrt(3)]/2

(b)

ok to make things simpler I will use a instead of theta

[(sin^2 a)/(1 - cos a)] = [(sec a + 1)/(sec a)]

Now do the Left Hand Side (LHS) first:

(sin^2 a)/(1 - cos a) = [(1 + cos a)(1 - cos a)]/(1 - cos a)

(using the property that sin^2 a = 1 - cos^2 a then using "difference of two squares" factorisation)

[(1 + cos a)(1 - cos a)]/(1 - cos a) ---> now cancel the like terms i.e. (1 - cos a)

Therefore LHS = 1 + cos a

Now for the Right Hand Side (RHS):

(sec a + 1)/(sec a) ---> split into fractions

=(sec a)/(sec a) +(1)/(sec a)

= 1 + 1/sec a

= 1 + cos a

Therefore LHS = RHS as required

Hope this helps!! :-)

Remind yourself of the equation for

cos(A-B)

That should crack the shell of the problem.