Question

tan-1 (√ 1-cosx / 1+cos), 0

Answer 1

Answer:

Step-by-step explanation:

∵ ($1-cosx=2sin^2(x/2)\\\\1+cosx=2cos^2(x/2)$)

$tan^{-1}(\sqrt{\dfrac{1-cosx}{1+cosx} } )\\\\\\=tan^{-1}(\sqrt{\dfrac{2sin^2(x/2)}{2cos^2(x/2)} } )\\\\\\=tan^{-1}(|tan(x/2)|)\\\\\\(0<x<\pi<=>0<\dfrac{x}{2} <\dfrac{\pi}{2} )\\\\\\=\dfrac{x}{2}$

Answer 2

Step-by-step explanation:

$\begin{lgathered}tan^{-1}(\sqrt{\dfrac{1-cosx}{1+cosx} } )\\\\\\=tan^{-1}(\sqrt{\dfrac{2sin^2(x/2)}{2cos^2(x/2)} } )\\\\\\=tan^{-1}(|tan(x/2)|)\\\\\\(00<\dfrac{x}{2} <\dfrac{\pi}{2} )\\\\\\=\dfrac{x}{2}\end{lgathered}$