Question

State whether the statement is true/false.

If ${\cos }^{-1}p+{\cos }^{-1}q+{\cos }^{-1}r=\pi$, then $p^2+q^2+r^2+2pqr=1$.

• A. True
• B. False

Answer 1

The correct option is A True

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

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

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

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

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

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

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

squaring both sides, we get

$p^2{q}^2+r^2+2pqr=(1-p^2)(1-q^2)$

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

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