Question

If ${\cos }^{-1}p+{\cos }^{-1}q+{\cos }^{-1}r=\pi$, then $p^2+q^2+r^2+2pqr$ is equal to:

• A. 3
• B. 1
• C. 2
• D. -1

Answer 1

The correct option is B 1

Given, ${\cos }^{-1}p+{\cos }^{-1}q+{\cos }^{-1}r=\pi$

$\Rightarrow {\cos }^{-1}(pq-\sqrt{(1-p^2)(1-q^2)})=\pi -{\cos }^{-1}r$

$\Rightarrow {\cos }^{-1}(pq-\sqrt{(1-p^2)(1-q^2)})=\frac{\pi }{2}+{\sin }^{-1}r$

$\Rightarrow {\sin }^{-1}(pq-\sqrt{(1-p^2)(1-q^2)})={\sin }^{-1}(-r)$

$\Rightarrow pq-\sqrt{(1-p^2)(1-q^2)}=-r$

$\Rightarrow pq+r=\sqrt{(1-p^2)(1-q^2)}$

$\Rightarrow p^2{q}^2+2pqr+r^2=1-p^2-q^2+p^2{q}^2$

$\Rightarrow p^2+q^2+r^2+2pqr=1$