Question

निम्नलिखित फलनों को सरलतम रूप में लिखिए:5. tan-1 √1+x2 -1 / x , x ≠ 0 ​.​

Answer 1

Answer:

the process of turning from liquid into vapour.

"snow cover prevents evaporation of water from the soil"

the process of something abstract ceasing to exist.

"thousands of employees witnessed the rapid evaporation of their retirement plans"

Answer 2

$\huge\mathtt{\fbox{\red{Answer}}}$

माना    $tan^{-1}x=\theta,i.e., \\\\x=tan\theta,-\dfrac{\pi}{2} < \theta< \dfrac{\pi}{2}$

अब  

        $tan^{-1}(\dfrac{\sqrt{1+x^2}-1 }{x})=tan^{-1}(\dfrac{\sqrt{1+tan^2\theta-1} }{tan\theta} ),\\ \\sec\theta>0..in (-\dfrac{\pi}{2} ,\dfrac{\pi}{2})\\\\\\=tan^{-1}(\dfrac{\sec\theta-1}{\tan\theta})\\ \\\\=tan^{-1}(\dfrac{1-cos\theta}{sin\theta} )\\\\\\=tan^{-1}(\dfrac{2sin^2(\theta/2)}{2sin(\theta/2) cos(\theta/2)} )\\\\\\=tan^{-1}(tan(\theta/2))\\\\=\dfrac{\theta}{2}\\ \\\\=\dfrac{1}{2} tan^{-1}x$

$|-\dfrac{\pi}{2} <\theta<\dfrac{\pi}{2}=-\dfrac{\pi}{4}<\dfrac{\theta}{2}<\dfrac{\pi}{4}$