Question

सिद्ध कीजिए कि ${\displaystyle {\tan }^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{(1+x^2)-\sqrt{1-x^2}}\right)=\frac{\pi }{2}+\frac{1}{2}{\cos }^{-1}{x}^2}$

Answer 1

L.H.S. ${\displaystyle {\tan }^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2-\sqrt{1-x^2}}}\right]}$

x² = cos θ रखें

∴ θ = ${\displaystyle {\cos }^{-1}{x}^2}$

⇒ ${\displaystyle {\tan }^{-1}\left[\frac{\sqrt{1+\cos \theta }+\sqrt{1-\cos \theta }}{\sqrt{1+\cos \theta }-\sqrt{1-\cos \theta }}\right]}$

⇒ ${\displaystyle {\tan }^{-1}\left[\frac{\sqrt{2{\cos }^2 \frac{\theta }{2}+\sqrt{2{\sin }^2 \frac{\theta }{2}}}}{\sqrt{2{\cos }^2 \frac{\theta }{2}-\sqrt{2{\sin }^2 \frac{\theta }{2}}}}\right]}$  ......${\displaystyle \left[\begin{array}{c}\text{क्योंकि} 1+\cos \theta =2{\cos }^2 \frac{\theta }{2}\\ 1-\cos \theta =2{\sin }^2 \frac{\theta }{2}\end{array}\right]}$

⇒ ${\displaystyle {\tan }^{-1}\left[\begin{array}{c}\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\\ \cos \frac{\theta }{2}-\sin \frac{\theta }{2}\end{array}\right]}$

⇒ ${\displaystyle {\tan }^{-1}\left[\begin{array}{c}1+\tan \frac{\theta }{2}\\ 1-\tan \frac{\theta }{2}\end{array}\right]}$  ......[ Nr. को विभाजित करना।और Den. द्वारा cos θ/2]

⇒ ${\displaystyle \tan \left[\tan (\frac{\pi }{4} \frac{\theta }{2})\right]}$  ......${\displaystyle [\text{क्योंकि}\frac{1+\tan \theta }{1-\tan \theta }=\tan (\frac{\pi }{4}+\theta )]}$

⇒ ${\displaystyle \frac{\pi }{4}+\frac{\theta }{2}}$

⇒ ${\displaystyle \frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}{x}^2}$ R.H.S. ......[θ = cos^–1x² लगाने पर]

इसलिए साबित हुआ।