Question #de713

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Question #de713

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Answer 1 · 2017-03-14T01:30:42

$L H S = tan 3 x$ $LHS=tan3x$

$= tan (x + 2 x)$ $=tan(x+2x)$

$= \frac{tan x + tan 2 x}{1 - tan x tan 2 x}$ $=(tanx+tan2x)/(1-tanxtan2x)$

$= \frac{tan x + \frac{2 tan x}{1 - {tan}^{2} x}}{1 - \frac{tan x \cdot 2 tan x}{1 - {tan}^{2} x}}$ $=(tanx+(2tanx)/(1-tan^2x))/(1-(tanx*2tanx)/(1-tan^2x)$

$= \frac{\frac{tan x (1 - {tan}^{2} x) + 2 tan x}{1 - {tan}^{2} x}}{\frac{1 - {tan}^{2} x - 2 {tan}^{2} x}{1 - {tan}^{2} x}}$ $=((tanx(1-tan^2x)+2tanx)/(1-tan^2x))/((1-tan^2x-2tan^2x)/(1-tan^2x))$

$= \frac{3 tan x - {tan}^{3} (x)}{1 - 3 {tan}^{2} (x)} = R H S$ $= (3tanx-tan^3(x))/(1-3tan^2(x))=RHS$

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$L H S = sin 4 x$ $LHS=sin4x$

$= 2 sin 2 x cos 2 x$ $=2sin2xcos2x$

$= 4 sin x cos x (1 - 2 {sin}^{2} x)$ $=4sinxcosx(1-2sin^2x)$

$= cos x \cdot 4 sin x (1 - 2 {sin}^{2} x)$ $=cosx*4sinx(1-2sin^2x)$

$= (cos x) (4 sin x - 8 {sin}^{3} x) = R H S$ $=(cosx)(4sinx-8sin^3x)=RHS$

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