Question

If tan(π/4 + (1/2) cos^-1x) + tan(π/4 - (1/2) cos^-1x) = 1, then x =
a) 0 b) 2 c) 3
d) No solution e) -2/3

Answer 1

$$\begin{gathered} \tan \left(\frac{\mathrm{\pi }}{4}+\frac{{\cos }^{-1}x}{2}\right)+\tan \left(\frac{\mathrm{\pi }}{4}-\frac{{\cos }^{-1}x}{2}\right)=1 \\ Use \\ \tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y}and\tan \left(x-y\right)=\frac{\tan x-\tan y}{1+\tan x\tan y} \\ \frac{\tan \frac{\mathrm{\pi }}{4}+\tan \left(\frac{{\cos }^{-1}x}{2}\right)}{1-\tan \frac{\mathrm{\pi }}{4}\tan \left(\frac{{\cos }^{-1}x}{2}\right)}+\frac{\tan \frac{\mathrm{\pi }}{4}-\tan \left(\frac{{\cos }^{-1}x}{2}\right)}{1+\tan \frac{\mathrm{\pi }}{4}\tan \left(\frac{{\cos }^{-1}x}{2}\right)}=1 \\ \frac{1+\tan \left(\frac{{\cos }^{-1}x}{2}\right)}{1-\tan \left(\frac{{\cos }^{-1}x}{2}\right)}+\frac{1-\tan \left(\frac{{\cos }^{-1}x}{2}\right)}{1+\tan \left(\frac{{\cos }^{-1}x}{2}\right)}=1 \\ Put\tan \left(\frac{{\cos }^{-1}x}{2}\right)=t \\ \frac{1+t}{1-t}+\frac{1-t}{1+t}=1 \\ \frac{{\left(1+t\right)}^2+{\left(1-t\right)}^2}{\left(1-t\right)\left(1+t\right)}=1 \\ \frac{\left(1+t^2+2t\right)+\left(1+t^2-2t\right)}{\left(1-t^2\right)}=1 \\ \frac{2+2t^2}{1-t^2}=1 \\ 2+2t^2=1-t^2 \\ 3t^2=-1 \\ Notpossible.Squareofarealnumberisalwaysnon-negative \\ Nosolution \end{gathered}$$