Question

d/dx tan⁻¹ √1-cosx/1+cosx=....... π < x < 2π,Select Proper option from the given options.
(a) 1/1+cos²x
(b) - 1/1+cos²x
(c) 1/2
(d) - 1/2

Answer 1

In the attachment I have answered this problem.         The given function is simplified in such a way that it is easy to find  derivative.               See the attachment for detailed solution.

Answer 2

you should first of all, solve $\sqrt{\frac{1-cosx}{1+cosx}}$

we know, $1-cos2\theta=2sin^2\theta$
$1+cos2\theta=2cos^2\theta$

so, $1-cosx=2sin^2\frac{x}{2}$

and $1+cosx=2cos^2\frac{x}{2}$

now, $\sqrt{\frac{1-cosx}{1+cosx}}=\sqrt{\frac{2sin^2\frac{x}{2}}{2cos^2\frac{x}{2}}}\\\\=\sqrt{tan^2\frac{x}{2}}\\\\=|tan\frac{x}{2}|$

but we know, x belongs to π < x < 2π
so, π/2 < x/2 < π , this means (x/2) belongs to 2nd quadrant . in 2nd quadrant tangent is negative. so, $|tan\frac{x}{2}|=-tan\frac{x}{2}$

now, we can write $\sqrt{\frac{1-cosx}{1+cosx}}=-tan\frac{x}{2}$

so, $\frac{d}{dx}tan^{-1}[-tan\frac{x}{2}]$

$=\frac{d}{dx}\{\frac{x}{2}\}$

[ because in π/2 < r < π, tan^-1(tanr) = -r ]

now, differentiate with respect to x,

so, you will get, $\frac{d}{dx}\{\frac{x}{2}\}$

henxez option (c) is correct.