Question 5011
instead of theta, i will use A..easier to write here


1. sinA(cotA+tanA) = secA

sinAcotA+sinAtanA = secA
sinAcotA+sinAtanA = 1/cosA
cosA(sinAcotA+sinAtanA) = 1
cosAsinAcotA + cosAsinAtanA = 1


right then...tanA = {{{sinA/cosA}}} and cotA = {{{cosA/sinA}}}, so:


{{{cosAsinAcosA/sinA + cosAsinAsinA/cosA = 1}}}
cosAcosA + sinAsinA = 1
{{{sin^2(A) + cos^2(A) = 1}}}


2. (sin theta + cos theta)^2 + (sin theta - cos theta)^2 = 2

(sinA + cosA)^2 + (sinA - cosA)^2 = 2
sin^2A + cos^2A + 2sinAcosA + sin^2A + cos^2A - 2sinAcosA = 2
2(sin^2A + cos^2A) = 2
sin^2A + cos^2A = 1


jon.