# Prove: `((1+1/tanx)(tan^2x - sin^2 x))/(tan^2x)=1` Thank you

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We have

`(1+1/tan(x))=(1+cos(x)/sin(x))`

`=(sin(x)+cos(x))/(sin(x))` (i)

`(tan^2(x)-sin^2(x))=((sin^2(x))/(cos^2(x))-sin^2(x))`

`=sin^2(x){1/(cos^2(x))-1}`

`=sin^2(x){(1-cos^2(x))/(cos^2(x))}`

`=(sin^4(x))/(cos^2(x))` (ii)

Thus

`{(1+1/(tan(x)))(tan^2(x)-sin^2(x))}/(tan^2(x))`

`={((sin(x)+cos(x))/(sin(x)))(sin^4(x))/(cos^2(x))}/(tan^2(x))`

`={(sin(x)+cos(x))tan^2(x)sin^2(x)}/{sin(x)tan^2(x)}`

`=(sin(x)+cos(x))sin(x)!=1`

May possible question not correct.

```"No true!"`

`((1+1/tanx)(tan^2x-sin^2 x))/(tan^2x)=1`

`(tanx+1)/tan x xx (tan^2 x -sin^2 x)/(tan^2 x) =1`

`(tan x +1 )/tan x xx (tan^2x- (tan^2x)/(1+tan^2 x))/(tan^2x)=1`

`(tan x+1)/(tan^3 x) xx (tan^2 x+tan^4 x-tan^2x)/(1+tan^2 x)= 1`

`(tan x +1)/(tan^3 x) xx (tan^4 x)/(1+tan^2x)=1`

`(tan x +1)(tan x)/(1+tan^2 x)=1`

`(tan^2 x + tan x)/(1+tan^2x)=1 `

`(tan^2 x +1+tanx -1)/(1+tan^2x)=1`

`1+ (tanx -1)/(1+tan^2x)=1`

`(tanx-1)/(1+tan^2x)=0`

then it does hold true only if tan x =1

Indeed i.e: `x= pi/6` `tan x= sqrt(3)/3` `sin x = 1/2`

`(1+sqrt(3))(1/3-1/4)/(1/3)=` `(1+sqrt(3)) xx 1/12 xx 3=` `(1+sqrt(3))/4 != 1`