Question

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

Answer 1

Answer:

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

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

अब  

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

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

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  

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

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