Prove the following-tan-4-x-tan-4-x-1-tan x-1-tan x-2

We have L.H.S. = tan ( π/4+x )/tan ( π/4 x ) tanπ/4+tan x/1 tanπ/4tan x/tanπ/4 tan x/1+tanπ/4tan x [ ∵tan(A+B)= tan A+tan B/1 tan Atan B; ∵tan(A B)= tan A ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

Prove the fol...

Question

Prove the following:

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

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Solution

We have
L.H.S. = $\frac{tan (\frac{π}{4} + x)}{tan (\frac{π}{4} - x)}$

$\frac{\frac{tan \frac{π}{4} + tan x}{1 - tan \frac{π}{4} tan x}}{\frac{tan \frac{π}{4} - tan x}{1 + tan \frac{π}{4} tan x}}$

$⎡ ⎢ ⎣ \begin{matrix} ∵ tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B} ∵ tan (A - B) = \frac{tan A - tan B}{1 - tan A tan B} \end{matrix} ⎤ ⎥ ⎦$

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

$= {[\frac{1 + tan x}{1 - tan x}]}^{2}$ = R.H.S

Suggest Corrections

Prove the following: tan(π4+x)tan(π4−x) = [1+tan x1−tan x]2

We have L.H.S. = tan(π4+x)tan(π4−x) tanπ4+tan x1−tanπ4tan xtanπ4−tan x1+tanπ4tan x ⎡⎢⎣∵tan(A+B)=tan A+tan B1−tan Atan B∵tan(A−B)=tan A−tan B1−tan Atan B⎤⎥⎦ 1+tan x1−tan x1−tan x1+tan x=(1+tan x)2(1−tan x)2 =[1+tan x1−tan x]2 = R.H.S

Prove the following:

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