2006-09-25 14:16:53 · 7 answers · asked by sadfjlhasldf 1 in Science & Mathematics ➔ Mathematics
Find the common denominator, which is sinXcosX
Then multiply each fraction, top AND bottom, by what its bottom is missing from the common denominator.
1/sinX times cosX/cosx + 1/cosX times sinX/sinX
This is now cosX/(sinXcosX) + sinX/(sinXcosX), or (cosX + sinx)/(sinXcosX)
2006-09-25 14:21:34 · answer #1 · answered by hayharbr 7 · 1⤊ 1⤋
solution for what?
what exactly do yo need?
you could write it as a single fraction:
1/sinX + 1/cosX = (cos x + sin x)/sin x cos x
2006-09-25 21:35:58 · answer #2 · answered by locuaz 7 · 0⤊ 0⤋
firstly the common denominator
is sinXcosX
then mutiply 1st part by the missing factor in the denominator and numerator
1/sinx+1/cosx
=1/sinX *( cosX/cosx) + 1/cosX *( sinX/sinX)
=cosX/(sinXcosX) + sinX/(sinXcosX)
=(cosX + sinx)/(sinXcosX) =answer
2006-09-26 01:50:10 · answer #3 · answered by KSA 3 · 0⤊ 0⤋
1/(sin x) + 1/(cos x)
=
(cos x)/(sin x cos x) + (sin x)/(sin x cos x)
=
(cos x + sin x) / (sin x cos x)
Time to throw in some fun math. First of all,
cos x + sin x = sqrt 2 * sin (x + 45*)
(Here 45* stands for 45 degrees; if you wanna be cool, write pi/4.)
Moreover,
sin x cos x = 1/2 sin (2x)
So you can simplify further as
1/sin x + 1/cos x = [sin (x+45*)] / [sqrt 2 * sin (2x)]
what is bound to amaze your teacher.
2006-09-26 01:16:14 · answer #4 · answered by dutch_prof 4 · 0⤊ 0⤋
cosx+sinx/cosxsinx
2006-09-26 08:39:22 · answer #5 · answered by ad2006miral 3 · 0⤊ 0⤋
(1/sin(x)) + (1/(cos(x))
csc(x) + sec(x)
2006-09-25 23:49:58 · answer #6 · answered by Sherman81 6 · 0⤊ 0⤋
find the common denominator, then add.
2006-09-25 21:58:27 · answer #7 · answered by davidosterberg1 6 · 0⤊ 1⤋