(secx-cosx) / (tanx-sinx) = cosecx+cotx

Spent hours to solve this question but failed. Please help. Thank you

1 answer

Unless I see an obvious identity I can replace, I usually change everything to sines and cosines.

LS = (1/cosx - cosx)/(sinx/cosx - sinx)
= ( 1 - cos^2 x)/cosx )/( (sinx - sinxcosx)/cosx )
= ( 1 - cos^2 x)/cosx )* cosx/( (sinx - sinxcosx)
= (1 - cos^2 x)/(sinx(1 - cosx))
= (1 - cosx)(1 + cosx)/(sinx(1 - cosx))
= (1 + cosx)/sinx
= 1/sinx + cosx/sinx
= cscx + cotx
+ RS
Similar Questions
  1. My previous question:Verify that (secx/sinx)*(cotx/cscx)=cscx is an identity. (secx/sinx)*(cotx/cscx) = (secx/cscx)(cotx/sinx) =
    1. answers icon 2 answers
  2. Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
    1. answers icon 0 answers
  3. 1/tanx-secx+ 1/tanx+secx=-2tanxso this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx)
    1. answers icon 0 answers
    1. answers icon 2 answers
more similar questions