How you prove [(sin^3x-csc^3x)/(sinx-cscx)]-(sinx-cscx)^2 ?

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Answer 1 · 2018-04-15T05:21:49

# 3#.

Knowing that, #a^3-b^3=(a-b)(a^2+ab+b^2)#, we have,

#(sin^3x-csc^3x)/(sinx-cscx)-(sinx-cscx)^2#,

#={cancel((sinx-cscx))(sin^2x+sinxcscx+csc^2x)}/cancel((sinx-cscx))#

#-(sin^2x-2sinxcscx+csc^2x)#,

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

#=1+2#,

#=3#.