prove that one side equals another.. please .. i am stuck
2006-12-08 12:48:02 · 14 answers · asked by lx88xl 1 in Science & Mathematics ➔ Mathematics
I'm going to assume you mean (cosx + 1) on the left.
(cosx + 1)/cotx
= (cosx + 1)/(cosx/sinx)
= (cosx + 1) * (sinx/cosx)
= sinx(cosx + 1)/cosx
= (sinxcosx + sinx)/cosx
= sinxcosx / cosx + sinx / cosx
= sinx + tanx
2006-12-08 12:54:03 · answer #1 · answered by Jim Burnell 6 · 3⤊ 0⤋
cos x+1/cot=sinx+tanx
you will modify side one to equal side two. First change the cotx into cosx/sinx:
cosx+1/ cosx/sinx= sinx + tanx
Now, flip the denominator and multiply it by the numberator, so you cancel the sins in the first term, and get cosx:
cosx+1/ (sinx/cosx)= sinx + tanx
Now your problem looks like this, so all you need to do is multiply the 1 by sinx/cosx, which is also called tanx, and as we know anything times 1 is the original term, so you are left with tanx:
sinx +1(sinx/cosx)= sinx + tanx
Now you are done, and you have:
sinx + tanx = sinx + tanx
2006-12-08 13:04:36 · answer #2 · answered by Christie 3 · 1⤊ 0⤋
ur question is written wrongly
as 1/cotx = tan x
so u r trying to say cosx = sinx and this can be
so I'll suppose u meant
RHS = (cosx + 1)/cotx
= cosx/cotx + 1/cotx
= cosx.sinx /cosx + tanx
= sinx + tanx = LHS....QED
2006-12-08 15:47:08 · answer #3 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
I'm going to assume you mean (cosx + 1) on the left.
(cosx + 1)/cotx
= (cosx + 1)/(cosx/sinx)
= (cosx + 1) * (sinx/cosx)
= sinx(cosx + 1)/cosx
= (sinxcosx + sinx)/cosx
= sinxcosx / cosx + sinx / cosx
= sinx + tanx
2006-12-08 13:19:22 · answer #4 · answered by Anonymous · 1⤊ 0⤋
(cosx+1)/cotx
=(cosx+1)tanx
=(cosx)tanx+tanx
=sinx+tanx
your question is not clear..as one may think even it as cosx+ 1/cotx
=cosx+tanx
which will not be equal to your right side part..
2006-12-08 16:27:28 · answer #5 · answered by bubbly 2 · 0⤊ 0⤋
I assume you mean (cosx + 1) on the left.
(cosx + 1)/cotx
= (cosx + 1)/(cosx/sinx)
= (cosx + 1) X (sinx/cosx)
= sinx(cosx + 1)/cosx
= (sinxcosx + sinx)/cosx
= sinxcosx / cosx + sinx / cosx
= sinx + tanx
2006-12-08 14:30:45 · answer #6 · answered by sanjay rock 1 · 1⤊ 0⤋
ur lhs should be (cosx + 1)/cot x
lhs: (Cosx + 1)/cot x = cosx / cot x + 1 / cotx
= (cosx.sinx)/cosx + tanx (since cotx = 1/tanx = cosx /sinx)
= sinx + tanx
= rhs
2006-12-08 16:24:15 · answer #7 · answered by mr_BIG 3 · 1⤊ 0⤋
cosx + 1)/cotx
= (cosx + 1)/(cosx/sinx)
= (cosx + 1) * (sinx/cosx)
= sinx(cosx + 1)/cosx
= (sinxcosx + sinx)/cosx
= sinxcosx / cosx + sinx / cosx
= sinx + tanx
2006-12-08 13:06:48 · answer #8 · answered by Aditya N 2 · 1⤊ 0⤋
cosx+1/cotx = (cosx+1)/(cosx/sinx)
= (cosxsinx+sinx)/cosx
= cosxsinx/cosx + sinx/cosx
=sinx + tanx
2006-12-09 00:48:46 · answer #9 · answered by raghunandan r 1 · 1⤊ 0⤋
I think you mean
(cos x + 1)/cot x = sin x + tan x
It is always helpful to convert each trig function into sin and cos.
(cos x + 1)/(cos x/sin x) = sin x + sin x/cos x
On the left side, multiply sin x/sin x. On the right, get the LCD
(sin x cos x + sin x)/cos x = (sin x cos x + sin x)/cos x
Therefore, since the expression on the left side is exactly the same as the expression on the right, then we can say that the identity is true, QED.
^_^
2006-12-08 13:03:10 · answer #10 · answered by kevin! 5 · 1⤊ 0⤋