The incetre of the triangle formed by the lines `x cos alpha+y sin alpha=pi, x cos beta+y sin beta=pi, x cos gamma+y sin gamma=pi` is `(h,k)` then `h+k=`

The orthocentre of triangle formed by (a cos alpha,a sin alpha),(a cos beta,a sin beta),(a cos gamma,a sin gamma) is

Find the area of the triangle whose vertices are : (a cos alpha, b sin alpha), (a cos beta, b sin beta), (a cos gamma, b sin gamma)

If x cos alpha + y sin alpha = x cos beta + y sin beta = 2a then cos alpha cos beta =

If (0,0) is orthocentre of triangle formed by A(cos alpha,sin alpha),B(cos beta,sin beta),C(cos gamma,sin gamma) then /_BAC=

Prove that the area of the parallelogram formed by the lines x cos alpha+y sin alpha=p,x cos alpha+y sin alpha=q,x cos beta+y sin beta=r and x cos beta+y sin beta=sis+-(p-q)(r-s)csc(alpha-beta)

If origin is the orthocenter of a triangle formed by the points (cos alpha*sin alpha,0)*(cos beta,sin beta.0),(cos gamma,sin gamma,0) then sum cos(2 alpha-beta-gamma)=

If (0,0) is orthocentre of triangle formed by A(cos alpha,sin alpha),B(cos beta,sin beta),C(cos gamma,sin gamma) then /_BAC is

If x cos alpha + y sin alpha = x cos beta + y sin beta = a, 0

If origin is the orthocentre of the triangle with vertices A(cos alpha,sin alpha),B(cos beta,sin beta),C(cos gamma,sin gamma) then cos(2 alpha-beta-gamma)+cos(2 beta-gamma-alpha)+cos(2 gamma-alpha-beta)=