Login

Class-12-science»Maths

tan inverse of x= sin inverse of (x/root of 1+xsquare) how?

$L e t \tan^{- 1} x = θ ....... (1) \Rightarrow \tan θ = x \Rightarrow \frac{\sin θ}{\cos θ} = x \Rightarrow \frac{\sin θ}{\sqrt{1 - \sin^{2} θ}} = x s q u a r i n g o n b o t h s i d e s, w e h a v e, \frac{\sin^{2} θ}{1 - \sin^{2} θ} = x^{2} \Rightarrow \sin^{2} θ = x^{2} (1 - \sin^{2} θ) \Rightarrow \sin^{2} θ + x^{2} \sin^{2} θ = x^{2} \Rightarrow \sin^{2} θ (1 + x^{2}) = x^{2} \Rightarrow \sin^{2} θ = \frac{x^{2}}{1 + x^{2}} \Rightarrow \sin θ = \frac{x}{\sqrt{1 + x^{2}}} \Rightarrow θ = \sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}}) ..... (2) F r o m (1) a n d (2), w e h a v e, \tan^{- 1} x = \sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}})$

1

View Full Answer

Copyright © 2024 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?