 if ${\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi ,{\text{ then }}{p^2} + {q^2} + {r^2} + 2pqr$ is equal to? (a)3 (b) 1 (c) 2 (d)-1
 if ${\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi ,{\text{ then }}{p^2} + {q^2} + {r^2} + 2pqr$ is equal to? (a)3 (b) 1 (c) 2 (d)-1
 hint- we will use the basic definitions of inverse cosine to solve this question.as we know, if $y = {\cos^{ - 1}}x$ then \[x\] must be in the range of (-1, 1) and y must be in the range of $(0,\pi ).$complete step-by-step solution -given equation is ${\cos ^{ - 1}}p + {\cos ^{ - 1}}q + {\cos ^{ - 1}}r = 3\pi $let ${y\_1} = {\cos ^{ - 1}}p,{y\_2} = {\cos ^{ - 1}}q,{y\_3} = {\cos ^{ - 1}}r$now, we have to evaluate the values of p, q, r.as we know if $y = {\cos^{ - 1}}x,{\text{ then - 1}} \leqslant {\text{x}} \leqslant {\text{1 and 0}} \leqslant {\text{y}} \leqslant \pi $hence, the given equation will hold only when each has the highest value of y.so, ${y\_1} = {y\_2} = {y\_3} = \pi $$   \rightarrow {\cos ^{ - 1}}p = {\cos ^{ - 1}}q = {\cos ^{ - 1}}r = \pi \\   \rightarrow p = q = r = \cos \pi \\   \rightarrow p = q = r = - 1 \\  $now, we have to find the value of ${p^2} + {q^2} + {r^2} + 2pqr$by substituting the value $p = q = r = - 1$ we get$   \rightarrow {( - 1)^2} + {( - 1)^2} + {( - 1)^2} + 2( - 1)( - 1)( - 1) \\   \rightarrow 3 - 2 \\   \rightarrow 1 \\  $hence, the correct option is b.note- to solve inverse trigonometric equations remember the basic concept of solving the algebraic equations and remember the definitions of inverse trigonometric functions such as about their domain and range. the identities of inverse trigonometric functions must be remembered with their domain and range.