tan(4π−x)tan(4π+x)=(1−tanx1+tanx)2.

use identitytan(A±B) =1∓tanAtanB tanA±tanB LHS=tan(4π −x) tan(4π +x) =1+tanx1−tanx 1−tanx1+tanx LHS=(1−tanx1+tanx )2=RHSHence proved

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{ ( {}{4} + x )}{ ( {}{4} - x )} = ( {1 + x}{1 - x} )^{2

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

11 mins ago

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11 mins ago

use identity

tan (A \pm B) = \frac{tan A \pm tan B}{1 \mp tan A tan B}

L H S = \frac{t an ( \frac{π}{4} + x )}{t an ( \frac{π}{4} - x )} = \frac{\frac{1 + t an x}{1 - t an x}}{\frac{1 - t an x}{1 + t an x}}

L H S = (\frac{1 + t an x}{1 - t an x})^{2} = R H S

Hence proved

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Question Text	$\frac{tan ( \frac{π}{4} + x )}{tan ( \frac{π}{4} - x )} = (\frac{1 + tan x}{1 - tan x})^{2}$ .
Answer Type	Text solution:1
Upvotes	150