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1/sinx - sinx = cosx/tanx
Simplify #3:
[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
1 answer
asked by
Anonymous
1,093 views
Q: If y=sinx/(1+tanx), find value of x not greater than pi, corresponding to maxima or minima value of y. I have proceeded thus-
5 answers
asked by
MS
1,017 views
Trigonometric Identities
Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
0 answers
asked by
Dave
1,462 views
tanx+secx=2cosx
(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
0 answers
asked by
shan
994 views
Prove the following identity:
1/tanx + tanx = 1/sinxcosx I can't seem to prove it. This is my work, I must've made a mistake
1 answer
asked by
Heather
808 views
Verify the identity:
tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx
0 answers
asked by
Ashley
1,271 views
Proving identity
(sinx+tanx)/(cosx+1)=tanx RS: (sinx+(sinx/cosx))/(cosx+1) ((sinxcosx/cosx)+(sinx/cosx))x 1/(cosx+1)
1 answer
asked by
sh
636 views
1/tanx-secx+ 1/tanx+secx=-2tanx
so this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx)
0 answers
asked by
olivia
1,353 views
Prove the following:
[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
3 answers
asked by
Anonymous
953 views
How do I solve this?
tan^2x= 2tanxsinx My work so far: tan^2x - 2tanxsinx=0 tanx(tanx - 2sinx)=0 Then the solutions are: TanX=0
1 answer
asked by
Chris
3,815 views