Tan -4 - x tan -4 - x - 1 - tan x1 - tan x 2-

Use identity tan( A± B ) #160; =tan A ±tan B #160;/ 1∓tan A tan B #160; LHS = tan( π #160; 4 +x ) #160;/ tan( π #160; 4 x ) #160; ...

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Trigonometric Ratios for Sum of Two Angles

tan π4 + x ta...

Question

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

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\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}

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

x \neq 0

tan (\frac{π}{4} + \frac{1}{2} {cos}^{- 1} x) + tan (\frac{π}{4} - \frac{1}{2} {cos}^{- 1} x), x \neq 0

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