The vertices of a triangle are `A(x_1, x_1, tan theta_1), B(x_2, x_2, tan theta_2)` and `C(x_3, x_3, tan theta_3)`. If the circumcentre coincides with origin then

The vertices of a triangle are : P(x_1,x_1 tan theta_1),Q (x_2, x_2tantheta_2) and R(x_3, x_3tantheta_3) . If the circumcentre of trianglePQR coincides with the origin and H(bar x,bar y) is the orthocentre, show that : bar y/bar x= (sintheta_1+sintheta_2+sintheta_3)/(costheta_1+costheta_2+costheta_3) .

The vertices of a triangle are A(x_(1),x_(1)tan theta_(1)),B(x_(2),x_(2)tan theta_(2))and C(x_(3),x_(3)tan theta_(3)) if the circumcentre of Delta ABC coincides with the origin and H(bar(x),bar(y)) is the orthocentre,show that (bar(y))/(bar(x))=(sin theta1+sin theta_(2)+sin theta_(3))/(cos theta_(1)+cos theta_(2)+cos theta_(3))

The vertices of a triangle are A(x_(1),x_(1)tan alpha),B(x_(2),x_(2)tan B eta) and C(x_(3),x_(3)tan gamma). If the circumcentre of Delta ABC coincides with the origin and H(a,b) be its orthocentre,then (a)/(b) is equal to

The vertices of a triangle lie on tlhe lines y=xtan theta_(1),y=x tan theta_(2),y=x tan theta_(3) . The circumcentre is the origin. Show that the locus of the orthocentre is x(sin theta_(1)+sin theta_(2)+sin theta_(3))=y(cos theta_(1)+cos theta_(2)+cos theta_(3)) .

tan theta=(2x(x+1))/(2x+1) then sin theta=

theta = tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) " then " tan theta=

Find, theta =tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) " then " tan theta=

theta = tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) " then " tan theta=

theta = tan^(-1) (2 tan^(2) theta) - tan^(-1) ((1)/(3) tan theta) " then " tan theta=

Prove that : tan 3 theta= (3 tan theta-tan^3 theta)/(1-3 tan^2 theta) .