Question

If $x>0,co{s}^{-1}(\frac{12}{x})=\frac{\pi }{2}-co{s}^{-1}(\frac{16}{x})$, then $x$ equals

• A. $12\times 16$
• B. $\frac{4}{3}$
• C. $\frac{3}{4}$
• D. $20$

Answer 1

Answer: (D)

$co{s}^{-1}\left(\frac{12}{x}\right)+co{s}^{-1}\left(\frac{16}{x}\right)=\frac{\pi }{2}$
$\Rightarrow \frac{12}{x}\cdot \frac{16}{x}-\sqrt{1-{\left(\frac{12}{x}\right)}^2}\sqrt{1-{\left(\frac{16}{x}\right)}^2}=cos\frac{\pi }{2}$
(using, $co{s}^{-1}x+co{s}^{-1}y=co{s}^{-1}(xy-\sqrt{(1-x^2)(1-y^2)}))$
$\Rightarrow \left[1-{\left(\frac{12}{x}\right)}^2\right]\left[1-{\left(\frac{16}{x}\right)}^2\right]={\left(\frac{12}{x}\right)}^2{\left(\frac{16}{x}\right)}^2$
$\Rightarrow {\left(\frac{12}{x}\right)}^2+{\left(\frac{16}{x}\right)}^2=1$
$\Rightarrow x=20$
Alternative Solution :
$co{s}^{-1}\left(\frac{12}{x}\right)=\frac{\pi }{2}-co{s}^{-1}\left(\frac{16}{x}\right)$
$\Rightarrow cos\left(co{s}^{-1}\left(\frac{12}{x}\right)\right)=cos\left(\frac{\pi }{2}-co{s}^{-1}\left(\frac{16}{x}\right)\right)$
$=cos\frac{\pi }{2}cos\left(co{s}^{-1}\left(\frac{16}{x}\right)\right)+sin\frac{\pi }{2}sin\left(co{s}^{-1}\left(\frac{16}{x}\right)\right)$
$\Rightarrow \frac{12}{x}=sinco{s}^{-1}\left(\frac{16}{x}\right)$
$\Rightarrow \frac{12}{x}=\sqrt{{\left(sinco{s}^{-1}\left(\frac{16}{x}\right)\right)}^2}$
$\Rightarrow {\left(\frac{12}{x}\right)}^2=1-{\left(cosco{s}^{-1}\left(\frac{16}{x}\right)\right)}^2$
$\Rightarrow {\left(\frac{12}{x}\right)}^2+{\left(\frac{16}{x}\right)}^2=1$
$\therefore x=20$
Short Cut Method :
Using fact : $co{s}^{-1}A+co{s}^{-1}B=\frac{\pi }{2}$ then
$A^2+B^2=1$
Now, $co{s}^{-1}(\frac{12}{x})+co{s}^{-1}(\frac{16}{x})=\frac{\pi }{2}$
$\Rightarrow (\frac{12}{x}{)}^2+(\frac{16}{x}{)}^2=1$
$\Rightarrow (12{)}^2+(16{)}^2=x^2$
$\Rightarrow x=20$ (as $x>0)$