Dfrac -tan left -dfrac -pi-4- - xright -tan left -dfrac -pi-4- - xright - - left -dfrac -1 - tan x-1 - tan x-right -2-

Question

Toppr Admin · Accepted Answer

Use identitytan-A-xB1-B-tanA-xB1-tanB1-x2213-tanAtanBLHS-tan-x3C0-4-x-tan-x3C0-4-x2212-x-1-tanx1-x2212-tanx1-x2212-tanx1-tanxLHS-1-tanx1-x2212-tanx-2-RHSHence proved

$\frac{tan (\frac{π}{4} + x)}{tan (\frac{π}{4} - x)} = {(\frac{1 + tan x}{1 - tan x})}^{2}$ .

use identity
$tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}$
$L H S = \frac{t a n (\frac{π}{4} + x)}{t a n (\frac{π}{4} - x)} = \frac{\frac{1 + t a n x}{1 - t a n x}}{\frac{1 - t a n x}{1 + t a n x}}$
$L H S = {(\frac{1 + t a n x}{1 - t a n x})}^{2} = R H S$
Hence proved

tan(π4+x)tan(π4−x)=(1+tanx1−tanx)2.

use identitytan(A±B)=tanA±tanB1∓tanAtanBLHS=tan(π4+x)tan(π4−x)=1+tanx1−tanx1−tanx1+tanxLHS=(1+tanx1−tanx)2=RHSHence proved

$\frac{tan (\frac{π}{4} + x)}{tan (\frac{π}{4} - x)} = {(\frac{1 + tan x}{1 - tan x})}^{2}$ .

use identity
$tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}$
$L H S = \frac{t a n (\frac{π}{4} + x)}{t a n (\frac{π}{4} - x)} = \frac{\frac{1 + t a n x}{1 - t a n x}}{\frac{1 - t a n x}{1 + t a n x}}$
$L H S = {(\frac{1 + t a n x}{1 - t a n x})}^{2} = R H S$
Hence proved