Question

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

Answer 1

`-1/2`

Explanation:

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

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

Explanation:

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`