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

Question

22600 views around the world

You can reuse this answer
Creative Commons License

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#