2007-01-08 01:17:38 · 4 answers · asked by sagar t 1 in Science & Mathematics ➔ Mathematics
One trick I use is to reduce all functions to a single one (or two)
cosec = 1/sin
cot = cos/sin
tan = sin/cos
1 = sin/sin = cos/cos = tan/tan = sin^2 + cos^2 = etc.
(cosecA+cotA+1)=
( (1/sin A) + (cosA/sinA) + (sinA/sinA)
((1 + cosA + sin A) / sinA)
2007-01-08 01:25:50 · answer #1 · answered by Raymond 7 · 0⤊ 0⤋
(cotA+cosecA-a million) = (cos A/sinA + a million/SinA -a million) = (cos A + a million - SinA )/Sin A (cotA - cosecA + a million) = (cos A/sinA - a million/SinA +a million) = (cos A - a million + SinA )/Sin A Divide = (cos A + a million - SinA )/(cos A - a million + SinA ) = [cos A + (a million - SinA )] / [cos A - (a million - SinA )] = [cos A + (a million - SinA )][cos A + (a million - SinA )] / [cos A - (a million - SinA )][cos A + (a million - SinA )] = [cos A + (a million - SinA )]^2 / [cos^2 A - (a million - SinA )^2] = (cos2A + a million + sin2 A - 2sin A + 2 cos A - 2 sin A cos A) = (a million + cos A )/sin A
2016-12-28 09:32:39 · answer #2 · answered by ? 3 · 0⤊ 0⤋
It's your homework right?
2007-01-08 01:18:43 · answer #3 · answered by Anonymous · 0⤊ 0⤋
this is your test right
2007-01-08 01:19:22 · answer #4 · answered by karthik r 1 · 0⤊ 0⤋