M2 trigonometry &_&

a) tan (π/2 + θ) + cot (π/2 + θ)

= tan [π - (π/2 - θ)] + cot [π - (π/2 - θ)]

= - tan (π/2 - θ) - cot (π/2 - θ)

= - cot θ - tan θ

= - cos θ/sin θ - sin θ/cos θ

= - (sin2 θ + cos2 θ)/(sin θ cos θ)

= -1/(sin θ cos θ)

tan (π + θ) + cot (π + θ) = tan θ + cot θ = 1/(sin θ cos θ)

b1) From a, we have tan (π/2 + θ) + cot (π/2 + θ) + tan (π + θ) + cot (π + θ) = 0

b2) Using similar approach, we have:

tan (3π/2 + θ) + cot (3π/2 + θ) + tan (2π + θ) + cot (2π + θ) = 0

So continuing with this approach, we have, the final answer:

tan (2009π/2 + θ) + cot (2009π/2 + θ)

= tan (1004π + π/2 + θ) + cot (1004π + π/2 + θ)

= tan (π/2 + θ) + cot (π/2 + θ)

= -1/(sin θ cos θ)

c) For n is odd, we have:

nπ/2 + π/4 in quad. II or IV, thus:

tan (nπ/2 + π/4) = -1 and cot (nπ/2 + π/4) = -1

Hence tan (nπ/2 + π/4) + cot (nπ/2 + π/4) = -2

For n is even, we have:

nπ/2 + π/4 in quad. I or III, thus:

tan (nπ/2 + π/4) = 1 and cot (nπ/2 + π/4) = 1

Hence tan (nπ/2 + π/4) + cot (nπ/2 + π/4) = 2

Thus the identity holds for any positive integer n.