Question

sec(Q+x)+sec(Q-x)=2secQ then prove cos^2Q=1+cosx

Answer 1

sec(Q + x) + sec(Q - x) = 2secQ
1/cos(Q + x) + 1/cos(Q - x) = 2/cosQ
{cos(Q - x) + cos(Q + x)}/cos(Q+x).cos(Q-x) = 2/cosQ
use the formula,
cosC + cosD = 2cos(C + D)/.cos(C-D)/2

{2cosQ.cosx}/{cos²Q - sin²x } = 2/cosQ
cos²Q.cosx = cos²Q - sin²x
cos²Q.cosx = cos²Q - sin²x
cos²Q.cosx - cos²Q = -sin²x { use sin²x = (1 - cos²Q)}
cos²Q{cosx - 1} = (cosx +1)(cosx -1)
cos²Q = (1 + cosx )
hence proved //