Question

If $co{s}^{-1}p+co{s}^{-1}q+co{s}^{-1}r=\pi$ then $p^2+q^2+r^2=$

• A. 1
  
• B. pqr
  
• C. 1+pqr
  
• D. 1-2pqr

Answer 1

The correct option is D

1-2pqr

$co{s}^{-1}p+co{s}^{-1}q+co{s}^{-1}r=\pi$
$co{s}^{-1}p+co{s}^{-1}q=\pi -co{s}^{-1}r$
$\Rightarrow co{s}^{-1}(pq-\sqrt{1-p^2}\sqrt{1-q^2})=co{s}^{-1}(-r)$
$\Rightarrow pq-\sqrt{1-p^2}\sqrt{1-q^2}=-r$
Squaring and simplyfying
$(pq+r{)}^2={\left(\sqrt{1-p^2}\sqrt{1-q^2}\right)}^2$
$\Rightarrow p^2{q}^2+r^2+2pqr=(1-p^2)(1-q^2)$
$\Rightarrow p^2{q}^2+r^2+2pqr=1-p^2-q^2+p^2{q}^2$
$\Rightarrow p^2+q^2+r^2=1-2pqr$