Login

If tan-1{root (1+x²) - root(1-x²)}/{ root (1+x²) +root (1-x²)} =theta , then prove that x² =sin 2 theta .

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

25

View Full Answer

Copyright © 2023 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?