Prove (sin x - csc x)^2 = sin^2x + cot^2x - 1.(sinx−cscx)2=sin2x+cot2x−1. Can anyone help me on this?

3 Answers
Dean R.
Apr 26, 2018

Show (sin x - csc x)^2 (sinx−cscx)2 = sin^2 x + cot^2 x - 1=sin2x+cot2x−1

Explanation:

(sin x - csc x)^2 (sinx−cscx)2

= (sin x - 1/sin x)^2 =(sinx−1sinx)2

= sin^2 x - 2 sin x (1/sinx) + 1/sin ^2 x =sin2x−2sinx(1sinx)+1sin2x

= sin^2 x - 2 + 1 / sin^2 x=sin2x−2+1sin2x

= sin^2 x - 1 + ( -1 + 1 / sin^2 x )=sin2x−1+(−1+1sin2x)

= sin^2 x + {1 - sin^2 x }/{ sin^2 x} - 1=sin2x+1−sin2xsin2x−1

= sin^2 x + cos^2 x / sin^2 x - 1 =sin2x+cos2xsin2x−1

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

Narad T.
Apr 26, 2018

Please see the proof below

Explanation:

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

Ratnaker Mehta
Apr 26, 2018

Kindly find a Proof in the Explanation.

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!

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