Question

Tanx+tan2x+tan3x=tanx.tan2x.tan3x

Answer 1

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$\huge\bold\red{{solution:-}}$

we have,${tan3x}=tan{(x+2)}$

$=\frac{tan x + tan 2x}{1-tanx.tan2x},$

$i.e,tan 3x=\frac{tanx+tan2x}{1-tanx. tan2x}.$

$Therefore,(1-tanx.tan2x)tan3x=$$tanx+tan2x$

or, $tan 3x-tanx.tan2x.tan3x=tanx+tan2x$

➪$tan3x=tanx+tan2x+tanx.tan2x$

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