Simplify $\frac{2 (tan x - cot x)}{{tan}^{2} x - {cot}^{2} x}$ $[2(tanx-cotx)]/(tan^2x-cot^2x)$ ?

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Answer 1 · 2017-01-15T07:46:45

$L H S = \frac{2 (tan x - cot x)}{{tan}^{2} x - {cot}^{2} x}$ $LHS=(2(tanx-cotx))/(tan^2x-cot^2x)$

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

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

$= \frac{2 tan x}{1 + {tan}^{2} x} = sin 2 x = R H S$ $=(2tanx)/(1+tan^2x)=sin2x=RHS$

Proved

Answer 2 · 2017-01-15T07:47:22

$\frac{2 (tan x - cot x)}{{tan}^{2} x - {cot}^{2} x} = sin 2 x$ $[2(tanx-cotx)]/(tan^2x-cot^2x)=sin2x$

Using the identity $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$ ,

we can split the term $({tan}^{2} x - {cot}^{2} x) = (tan x + cot x) (tan x - cot x)$ $(tan^2x-cot^2x)=(tanx+cotx)(tanx-cotx)$ and we get

$\frac{2 (tan x - cot x)}{{tan}^{2} x - {cot}^{2} x}$ $[2(tanx-cotx)]/(tan^2x-cot^2x)$

= $\frac{2 (tan x - cot x)}{(tan x + cot x) (tan x - cot x)}$ $[2(tanx-cotx)]/((tanx+cotx)(tanx-cotx))$

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

= $\frac{2}{(\frac{sin x}{cos x} + \frac{cos x}{sin x})}$ $2/((sinx/cosx+cosx/sinx))$

= $\frac{2}{\frac{{sin}^{2} x + {cos}^{2} x}{sin x cos x}}$ $2/((sin^2x+cos^2x)/(sinxcosx)$

= $\frac{2}{\frac{1}{sin x cos x}}$ $2/(1/(sinxcosx)$

= $2 \times \frac{sin x cos x}{1}$ $2xx(sinxcosx)/1$

= $2 sin x cos x$ $2sinxcosx$

= $sin 2 x$ $sin2x$

Simplify [2(tanx-cotx)]/(tan^2x-cot^2x)2(tanx−cotx)tan2x−cot2x[2(tanx-cotx)]/(tan^2x-cot^2x)?