Prove #(sin x - csc x)^2 = sin^2x + cot^2x - 1.# Can anyone help me on this?

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Answer 1 · 2018-04-26T17:20:16

Show # (sin x - csc x)^2 ## = sin^2 x + cot^2 x - 1#

# (sin x - csc x)^2 #

#= (sin x - 1/sin x)^2 #

# = sin^2 x - 2 sin x (1/sinx) + 1/sin ^2 x #

# = sin^2 x - 2 + 1 / sin^2 x#

# = sin^2 x - 1 + ( -1 + 1 / sin^2 x )#

# = sin^2 x + {1 - sin^2 x }/{ sin^2 x} - 1#

# = sin^2 x + cos^2 x / sin^2 x - 1 #

# = sin^2 x + cot^2 x - 1 quad sqrt #

Answer 2 · 2018-04-26T17:22:52

Please see the proof below

We need

#cscx=1/sinx#

#sin^2x+cos^2x=1#

#1/sin^2x=1+cot^2x#

Therefore,

#LHS=(sinx-cscx)^2#

#=(sinx-1/sinx)^2#

#=sin^2x-2+1/sin^2x#

#=sin^2x-2+1+cot^2x#

#=sin^2x+cot^2x-1#

#=RHS#

#QED#

Answer 3 · 2018-04-26T17:23:09

Kindly find a Proof in the Explanation.

We will use the Identity : #cosec^2x=cot^2x+1#.

#(sinx-cosecx)^2#,

#=sin^2x-2sinx*cosecx+cosec^2x#,

#=sin^2x-2sinx*1/sinx+cot^2x+1#,

#=sin^2x-2+cot^2x+1#,

#=sin^2x+cot^2x-1#, as desired!