1/sinx - sinx = cosx/tanx

  1. Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
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    2. Anonymous asked by Anonymous
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  2. 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-
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    2. MS asked by MS
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  3. Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
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    2. Dave asked by Dave
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  4. tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
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    2. shan asked by shan
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  5. 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
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    2. Heather asked by Heather
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  6. Verify the identity:tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx
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    2. Ashley asked by Ashley
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  7. Proving identity(sinx+tanx)/(cosx+1)=tanx RS: (sinx+(sinx/cosx))/(cosx+1) ((sinxcosx/cosx)+(sinx/cosx))x 1/(cosx+1)
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    2. sh asked by sh
    3. views icon 636 views
  8. 1/tanx-secx+ 1/tanx+secx=-2tanxso this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx)
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    2. olivia asked by olivia
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  9. Prove the following:[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
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    2. Anonymous asked by Anonymous
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  10. 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
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    2. Chris asked by Chris
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