Question

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

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

Answer 1

Answer: (D)

$cos^{-1} \left(\frac{12}{x}\right) + cos^{-1} \left(\frac{16}{x}\right) = \frac{\pi}{2}$
$\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, $cos^{-1} x + cos^{-1} y = cos^{-1} (xy - \sqrt{(1 - x^2)(1 - y^2)} \,))$
$\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 :
$cos^{-1}\left(\frac{12}{x}\right) = \frac{\pi}{2} - cos^{-1} \left(\frac{16}{x}\right)$
$\Rightarrow cos \left(cos^{-1}\left(\frac{12}{x}\right)\right) = cos\left(\frac{\pi}{2} - cos^{-1}\left(\frac{16}{x}\right)\right) $
$= cos \frac{\pi}{2} cos\left(cos^{-1} \left(\frac{16}{x}\right)\right)+sin \frac{\pi}{2}sin\left(cos^{-1}\left(\frac{16}{x}\right)\right)$
$\Rightarrow \frac{12}{x} = sin \,cos^{-1}\left(\frac{16}{x}\right) $
$\Rightarrow \frac{12}{x} = \sqrt{\left(sin \,cos^{-1}\left(\frac{16}{x}\right)\right)^{2}} $
$\Rightarrow \left(\frac{12}{x}\right)^{2} = 1 - \left(cos \,cos^{-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 : $cos^{-1} A + cos^{-1} B = \frac{\pi}{2}$ then
$A^2 + B^2 = 1$
Now, $cos^{-1}(\frac{12}{x}) + cos^{-1}(\frac{16}{x}) = \frac{\pi}{2}$
$\Rightarrow (\frac{12}{x})^2 + (\frac{16}{x})^2 = 1$
$\Rightarrow (12)^2 + (16)^2 = x^2$
$\Rightarrow x = 20 $ (as $x > 0)$