Question

Aratati ca:
A)tgx+ctgx=2/sin2x
B)tg^2x+ct^2x=4(sin2x)^2-2

Answer 1

a)tgx+ctgx=2/2sinxcosx
tgx+ctgx=1/sinxcosx
sinx/cosx+cosx/sinx=1/sinxcosx
sin↑2x+cos↑2x=1(A)
b)sin↑2x/cos↑2x+cos↑2x/sin↑2x=4*4sin↑2xcos↑2x-2
sin↑4x+cos↑4x=16sin↑4xcos↑4x-2sin↑2xcos↑2x
$( sin^{2}x+ cos^{2}x) ^{2} =16 sin^{4} xcos^{4} x \\ 1=16sin^{4} xcos^{4} x$