Inverse Trigonometric Functions Question 57
Question: If $ {{\cos }^{-1}}p+{{\cos }^{-1}}q+{{\cos }^{-1}}r=\pi $ then $ p^{2}+q^{2}+r^{2}+2pqr= $
[Karnataka CET 2004]
Options:
A) 3
B) 1
C) 2
D) -1
Show Answer
Answer:
Correct Answer: B
Solution:
Trick: According to given condition, we put $ p=q=r=\frac{1}{2} $ .
Then, $ p^{2}+q^{2}+r^{2}+2pqr $ = $ {{( \frac{1}{2} )}^{2}}+{{( \frac{1}{2} )}^{2}}+{{( \frac{1}{2} )}^{2}}+2.\frac{1}{2}.\frac{1}{2}.\frac{1}{2} $ = $ \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{2}{8} $ =1.
BETA
