d/dx(tan^-1(cosx/1+sinx))?

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Answer 1 · 2018-06-23T08:52:17

$-1/2$

We bget

$1/(1+cos^2(x)/(1+sin(x))^2)*(-sin(x)(1+sin(x))-cos^2(x))/(1+sin(x))^2$
This is

$((1+sin(x))^2(-1-sin(x)))/(2(1+sin(x))(1+sin(x))^2)$
$=-1/2$

Answer 2 · 2018-06-23T09:09:31

$d/(dx)(tan^-1(cosx/(1+sinx)))=-1/2$

We know that,

$color(red)((1)sin(pi/2-theta)=costheta and cos(pi/2-theta)=sintheta$

$color(blue)((2)sin2alpha=2sinalpha cosalpha and 1+cos2alpha=2cos^2alpha$

Using (1) we get

$y=tan^-1(cosx/(1+sinx))=tan^-1(color(red)(sin(pi/2-x))/(1+color(red)(cos(pi/2-x))))$

For simplicity we take , $color(violet)(pi/2-x=2alpha=>alpha=pi/4-x/2$

So, $y=tan^-1((sin2alpha)/(1+cos2alpha))...tocolor(blue)(Apply(2)$

$=>y=tan^-1((2sinalphacosalpha)/(2cos^2alpha))$

$=>y=tan^-1(sinalpha/cosalpha)$

$=>y=tan^-1(tanalpha) ,where ,color(violet)( alpha=pi/4-x/2$

$=>y=alpha$

$=>y=pi/4-x/2$

Diff.w.r.t. $x$

$(dy)/(dx)=0-1/2=-1/2$