Refer to the following diagram :

`{:(,"Column I",,,"Column II"),(a.,"Collinear vectors",,p.,veca),(b.,"Coinitial vectors",,q.,vecb),(c.,"Equal vectors",,r.,vecc),(d.,"Unlike vectors (same initial point)",,s.,vecd):}`

Answer

2

Explanation

(q) `cos x + cos y + cosz =0 and sin x + siny + sinz =0`
`" " cos x + cosy = - cos z" " …(1)`
and `sin x + sin y = - sin z" " … (2)`
Squaring and adding we get,
` 1+1 +2 ( cos x cosy + sin x sin y ) =1`
`rArr 2+2 cos (x-y)=1`
`rArr 2 cos (x-y) =-1`
`rArr cos (x-y)=-(1)/(2)`
`rArr 2 cos^(2)((x-y)/(2)) -1 = -(1)/(2)`
`rArr cos ((x-y)/(2)) = (1)/(2)`
(r) `cos ((pi)/(4) - x) cos 2x + sin x sin 2x sec x`
`" " = cos x sin 2x secx + cos ((pi)/(4) +x) cos 2x`
`rArr [ cos ((pi)/(4) -x) - cos ((pi)/(4) + x)] cos 2x`
`" " = ( cos x sin 2x secx - sin x sin 2x secx)`
`rArr (2)/(sqrt2) sin x cos 2x = ( cos x - sin x ) sin 2x secx`
`rArr sqrt2 sinx cos 2x = (cosx - sin x) 2 sin x `
`rArr (1)/(sqrt2) = (1)/( cos x + sin x)`
`rArr x = (pi)/(4)`
`rArr secx = sec ""(pi)/(4) = sqrt2`