Question

tan-1 x/√a2 - x2 , |x| < a

Answer 1

Answer:

ok then what is the question

Answer 2

Answer:

Step-by-step explanation:

माना   $sin^{-1}(\dfrac{x}{a} )=\theta\\\\\\\dfrac{x}{a} =sin\theta,-\dfrac{\pi}{2} \leq \theta\leq \dfrac{\pi}{2}$

$tan^{-1}(\dfrac{x}{\sqrt{a^2-x^2} } )\\\\\\=tan^{-1}(\dfrac{asin\theta}{\sqrt{a^2-a^2sin^2\theta} } )\\\\\\=tan^{-1}(\dfrac{sin\theta}{\sqrt{1-sin^2\theta} } ),\\\\\\(cos\theta>0 for -\dfrac{\pi}{2} \leq \theta\leq \dfrac{\pi}{2} )\\\\\\=tan^{-1}(\dfrac{sin\theta}{cos\theta} )\\\\\\\=tan^{-1}(tan\theta)\\\\\\=\theta\\\\\\=sin^{-1}(\dfrac{x}{a} )\\\\\\(a>0)$