If ` cos 20^(@) - sin 20^(@) =p " then " cos 40^(@) ` is equal to

Options

A. `p^(2) sqrt(2-p^(2))`
B. `p sqrt(2-p^(2))`
C. `p+ sqrt(2-p^(2))`
D. ` p -sqrt(2-p^(2))`

Answer

B

Explanation

Given ` cos 20^(@) - sin 20^(@) =p`
On squaring both sides, we get
` cos^(2) 20^(@) + sin^(2) 20^(@) -2 sin 20^(@) cos 20^(@) =p^(2)`
` Rightarrow 1-p^(2) = sin 40^(@)`
` Rightarrow 1-p^(2) =sqrt(1-cos^(2) 40^(@))`
` Rightarrow (1-p^(2))=1 - cos^(2) 40^(@)`
` cos^(2) 40^(@) =1 -(1+p^(4) -2p^(2))`
` cos 40^(@) = sqrt(2p^(2) -p^(2))`
` cos 40^(@) =p sqrt(2-p^(2))`