Question

How to do the 19th question


$19.Findthevalueofcot\frac{1}{2}({\cos }^{-1}\frac{2x}{1+x^2}+{\sin }^{-1}\frac{1-y^2}{1+y^2}),\left|x\right|<1,y>0,xy<1$

Answer 1

$$\begin{gathered} cot\frac{1}{2}\left({\cos }^{-1}\frac{2x}{1+x^2}+{\sin }^{-1}\frac{1-y^2}{1-y^2}\right) \\ Conditioned:\left|x\right|<1,y>0,xy<1 \\ Thismeans \\ 0<y\le 1 \\ -1<x<1 \\ Letx=\tan \alpha ,\alpha \in \left(-\frac{\pi }{4},\frac{\pi }{4}\right)andy=\tan \beta ,\beta \in (0,\frac{\pi }{4}] \\ cot\frac{1}{2}\left({\cos }^{-1}\frac{2x}{1+x^2}+{\sin }^{-1}\frac{1-y^2}{1+y^2}\right)=cot\frac{1}{2}\left({\cos }^{-1}\frac{2\tan \alpha }{1+{\tan }^2\alpha }+{\sin }^{-1}\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }\right) \\ =cot\frac{1}{2}\left({\cos }^{-1}\left(sin2\alpha \right)+{\sin }^{-1}\left(\cos 2\beta \right)\right) \\ =cot\frac{1}{2}\left(\frac{\pi }{2}-{\sin }^{-1}\left(sin2\alpha \right)+\frac{\pi }{2}-{\cos }^{-1}\left(\cos 2\beta \right)\right) \\ =cot\frac{1}{2}\left(\pi -\left({\sin }^{-1}\left(\sin 2\alpha \right)+{\cos }^{-1}\left(\cos 2\beta \right)\right)\right) \\ Since2\alpha \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)and2\beta \in (0,\frac{\pi }{2}] \\ Therefore,{\sin }^{-1}\left(\sin 2\alpha \right)=2\alpha and{\cos }^{-1}\left(\cos 2\beta \right)=2\beta \\ =cot\frac{1}{2}\left(\pi -\left(2\alpha +2\beta \right)\right) \\ =cot\left(\frac{\pi }{2}-\left(\alpha +\beta \right)\right) \\ =\tan \left(\alpha +\beta \right) \\ \left\{Since\tan \alpha =xand\tan \beta =y,then\alpha ={\tan }^{-1}xand\beta ={\tan }^{-1}y\right\} \\ =\tan \left({\tan }^{-1}x+{\tan }^{-1}y\right) \\ =\tan \left({\tan }^{-1}\left(\frac{x+y}{1-xy}\right)\right) \\ =\frac{x+y}{1-xy} \end{gathered}$$