Prove that tan-4-x tan-4-x - 1- tan x1- tan x2

L.H.S = tan(π4+x )tan(π4 x ) tan (A ± B) =tan A±tan B1 ∓tan A·tan B = tanπ4+ tan x1 tanπ4tan xtanπ4 tan x1+ tanπ4tan x=1+ tan x1 tan x1 tan x1+ tan x =1+ ta ...

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Standard XII

Mathematics

Trigonometric Ratios of Common Angles

Prove that ta...

Question

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

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Solution

$L . H . S = \frac{tan (\frac{π}{4} + x)}{tan (\frac{π}{4} - x)}$
$tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A \cdot tan B}$
$= \frac{\frac{tan \frac{π}{4} + tan x}{1 - tan \frac{π}{4} tan x}}{\frac{tan \frac{π}{4} - tan x}{1 + tan \frac{π}{4} tan x}} = \frac{\frac{1 + tan x}{1 - tan x}}{\frac{1 - tan x}{1 + tan x}}$
$= \frac{1 + tan x}{1 - tan x} \times \frac{1 + tan x}{1 - tan x} = {(\frac{1 + tan x}{1 - tan x})}^{2}$
L.H.S. = R.H.S.
Hence proved.

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

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Trigonometric Ratios of Common Angles

Standard XII Mathematics

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Prove that tan(π4+x)tan(π4−x)=(1+tanx1−tanx)2

L.H.S=tan(π4+x)tan(π4−x) tan(A±B)=tanA±tanB1∓tanA⋅tanB =tanπ4+tanx1−tanπ4tanxtanπ4−tanx1+tanπ4tanx=1+tanx1−tanx1−tanx1+tanx =1+tanx1−tanx×1+tanx1−tanx=(1+tanx1−tanx)2 L.H.S. = R.H.S. Hence proved.

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