Simplify #cot^2x/(1-sinx)#?

1 Answer
Shwetank Mauria
Jan 3, 2018

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

Explanation:

#cot^2x/(1-sinx)#

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

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

= #(1-sin^2x)/sin^2x xx1/(1-sinx)#

= #((1-sinx)(1+sinx))/sin^2x xx1/(1-sinx)#

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

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

= #csc^2x+cscx#

= #cscx(1+cscx)#

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