ABC is a triangle. D is the middle point of BC. If AD is perpendicular to AC, The value of cos A cos C , is

Options

A. `(3(c^(2)-a^(2))/(ac)`
B. `(a^(2)-c^(2))/(2ac)`
C. `(2(c^(2)-a^(2))/(3ac)`
D. none of these

Explanation

We have,
`a^(2)-c^(2)=3b^(2)` and `cosC=(2b)/a`
`thereforecosAcosC=(b^(2)+c^(2)-a^(2))/(2bc)xx(2b)/a`
`rArrcosAcosC=((b^(2)+c^(2)-a^(2)))/(ac)`
`rArrcosAcosC=((a^(2)-c^(2))/3+c^(2)-a^(2))/(ac)[becauseb^(2)=(a^(2)-c^(2))/3]`
`rArrcosAcosC=(2(c^(2)-a^(2)))/(3ac)`