Question

Solve the following for $x:si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}$

OR

Show that: $2si{n}^{-1}\left(\frac{3}{5}\right)-ta{n}^{-1}\left(\frac{17}{31}\right)=\frac{\pi }{4}$

Answer 1

$si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}\Rightarrow si{n}^{-1}(1-x)=\frac{\pi }{2}+2si{n}^{-1}x\Rightarrow (1-x)=sin\left(\frac{\pi }{2}\right)+2si{n}^{-1}x\Rightarrow (1-x)=cos\left(2si{n}^{-1}x\right)\Rightarrow (1-x)=cos\left(co{s}^{-1}\right(1-2x^2\left)\right)\Rightarrow 1-x=1=2x^2\Rightarrow 2x^2-x=0\therefore x=0,x=\frac{1}{2}$

OR
$2si{n}^{-1}\left(\frac{3}{5}\right)-ta{n}^{-1}\left(\frac{17}{31}\right)=\frac{\pi }{4}L.H.S.,=co{s}^{-1}(1-2\times \frac{9}{25})-ta{n}^{-1}\left(\frac{17}{31}\right)=co{s}^{-1}\left(\frac{7}{25}\right)-ta{n}^{-1}\left(\frac{17}{31}\right)=ta{n}^{-1}\left(\frac{24}{7}\right)ta{n}^{-1}\left(\frac{17}{31}\right)=ta{n}^{-1}\left(\frac{24}{7}\right)ta{n}^{-1}\left(\frac{17}{31}\right)=ta{n}^{-1}\left(\frac{\frac{24}{7}-\frac{17}{31}}{1+\frac{24}{7}\times \frac{17}{31}}\right)=ta{n}^{-1}\left(\frac{24\times 31-17\times 7}{31\times 7+24\times 17}\right)=ta{n}^{-1}\left(\frac{625}{625}\right)=ta{n}^{-1}1=\frac{\pi }{4}=R.H.S.$

Hence Proved