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

[ L.H.S=tan #160; ( π #160; 4 +x ) #160; tan #160; ( π #160; 4 x ) #160; ...

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Sin(A+B)Sin(A-B)

Prove: tanπ...

Question

Prove: $\frac{t a n (\frac{π}{4} + x)}{t a n (\frac{π}{4} - x)} = (\frac{1 + t a n x}{1 - t a n x})^{2}$

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

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

Prove the following:

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

tan (π / 4 + θ) \times tan (π / 4 - θ) = 1

\frac{1 + s i n 2 x}{c o s 2 x} = t a n (\frac{π}{4} + x)

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