Question #4aae2

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Question #4aae2

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Answer 1 · 2017-02-12T15:03:35

$y = \frac{sin 4 x + cos 4 x}{sin 4 x - cos 4 x}$ $y=(Sin4x+cos4x)/(sin4x-cos4x)$

$\Rightarrow \frac{y}{1} = \frac{sin 4 x + cos 4 x}{sin 4 x - cos 4 x}$ $=>y/1=(sin4x+cos4x)/(sin4x-cos4x)$

By componendo and dividendo

$\Rightarrow \frac{y + 1}{y - 1} = \frac{2 sin 4 x}{2 cos 4 x}$ $=>(y+1)/(y-1)=(2sin4x)/(2cos4x)$

$\Rightarrow \frac{y + 1}{y - 1} = tan 4 x$ $=>(y+1)/(y-1)=tan4x$

$\Rightarrow \frac{y + 1}{y - 1} = \frac{2 tan 2 x}{1 - {tan}^{2} 2 x}$ $=>(y+1)/(y-1)=(2tan2x)/(1-tan^2 2x$

Again by componendo and dividendo

$\Rightarrow \frac{2 y}{2} = \frac{2 tan 2 x + 1 - {tan}^{2} 2 x}{2 tan 2 x - 1 + {tan}^{2} 2 x}$ $=>(2y)/2=(2tan2x+1-tan^2 2x)/(2tan2x-1+tan^2 2x)$

$\Rightarrow y = \frac{\frac{4 tan x}{1 - {tan}^{2} x} + 1 - \frac{4 {tan}^{2} x}{{(1 - {tan}^{2} x)}^{2}}}{\frac{4 tan x}{1 - {tan}^{2} x} - 1 + \frac{4 {tan}^{2} x}{{(1 - {tan}^{2} x)}^{2}}}$ $=>y=((4tanx)/(1-tan^2x)+1-(4tan^2x)/(1-tan^2x)^2)/((4tanx)/(1-tan^2x)-1+(4tan^2x)/(1-tan^2x)^2)$

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Question #4aae2

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