How to prove (tan^3x-cot^3x)/(tan^2x+csc^2x)=tanx-cotx?

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Answer 1 · 2018-04-29T15:59:34

Please see below.

We know that,

$(1) a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $color(red)((1)a^3-b^3=(a-b)(a^2+ab+b^2)$

$(2) {csc}^{2} x = 1 + {cot}^{2} x$ $color(blue)((2)csc^2x=1+cot^2x$

$(3) tan x cot x = 1$ $color(violet)((3)tanxcotx=1$

Here,

$\frac{{tan}^{3} x - {cot}^{3} x}{{tan}^{2} x + {csc}^{2} x} = tan x - cot x$ $(tan^3x-cot^3x)/(tan^2x+csc^2x)=tanx-cotx$

Let,

$LHS=color(red)((tan^3x-cot^3x))/(tan^2x+color(blue)(csc^2x))...toApply(1)and(2)$

$=color(red)(((tanx-cotx)(tan^2x+tanxcotx+cot^2x)))/(tan^2x+color(blue)(1+cot^2x))$

$=((tanx-cotx)(tan^2x+color(violet)(1)+cot^2x))/((tan^2x+1+cot^2x))...toApply(3)$

$=tanx-cotx$

$=RHS$