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Motion In A Plane

Question
CBSEENPH11018059

Show that:

Solution

We have the inequality, 
                 negative 1 less-than or slanted equal to cosθ less-than or slanted equal to 1  
i.e., negative 1 space less-than or slanted equal to space minus cos space straight theta space less-than or slanted equal to space 1 space

Now, multiplying the inequality by 2AB, 
negative 2 A B space less-than or slanted equal to space minus 2 A B space cos theta space less-than or slanted equal to space 2 A B space
Now adding A2 + B2, we get 
straight A squared space plus space straight B squared space minus space 2 AB space less-than or slanted equal to space straight A squared plus space straight B squared space minus space 2 ABcosθ

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space less-than or slanted equal to space straight A squared plus space straight B squared space plus space 2 AB

open parentheses open vertical bar open vertical bar straight A with rightwards harpoon with barb upwards on top close vertical bar minus open vertical bar straight B with rightwards harpoon with barb upwards on top close vertical bar close vertical bar close parentheses squared space less-than or slanted equal to space open parentheses open vertical bar straight A with rightwards harpoon with barb upwards on top minus straight B with rightwards harpoon with barb upwards on top close vertical bar close parentheses squared space less-than or slanted equal to space open parentheses open vertical bar straight A with rightwards harpoon with barb upwards on top close vertical bar plus open vertical bar straight B with rightwards harpoon with barb upwards on top close vertical bar close parentheses squared space
Taking square root, we get
open vertical bar open vertical bar straight A with rightwards harpoon with barb upwards on top close vertical bar minus open vertical bar straight B with rightwards harpoon with barb upwards on top close vertical bar close vertical bar space less-than or slanted equal to space open vertical bar A with rightwards harpoon with barb upwards on top minus B with rightwards harpoon with barb upwards on top close vertical bar space less-than or slanted equal to space open vertical bar A with rightwards harpoon with barb upwards on top close vertical bar space plus space open vertical bar B with rightwards harpoon with barb upwards on top close vertical bar
Hence, proved.