Question

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

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

Answer 1

The correct option is D $20$

Given for $x>0,{\cos }^{-1}\left(\frac{12}{x}\right)=\frac{\pi }{2}-{\cos }^{-1}\left(\frac{16}{x}\right)$
$\Rightarrow {\cos }^{-1}\left(\frac{12}{x}\right)+{\cos }^{-1}\left(\frac{16}{x}\right)=\frac{\pi }{2}$
$\Rightarrow {\cos }^{-1}\frac{12}{x}\cdot \frac{16}{x}-{\sqrt{1-\left(\frac{12}{x}\right)}}^2{\sqrt{1-\left(\frac{16}{x}\right)}}^2=\frac{\pi }{2}$
using ${\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}={\cos }^{-1}(xy-\sqrt{(1-x^2)(1-y^2)})$
$\Rightarrow \frac{12}{x}\cdot \frac{16}{x}-\sqrt{1-{\left(\frac{12}{x}\right)}^2}\sqrt{1-{\left(\frac{16}{x}\right)}^2}=\cos \frac{\pi }{2}$
$\Rightarrow {[1-\frac{12}{x}]}^2{[1-\frac{16}{x}]}^2={\left(\frac{12}{x}\right)}^2{\left(\frac{16}{x}\right)}^2$
$\Rightarrow x=20$