`(b^(2)-c^(2)) tan B tan C+(c^(2)-a^(2)) tan C tan A+(a^(2)-b^(2)) tan A tan B=0`

In a triangle ABC , (a^2-b^2-c^2) tan A +(a^2-b^2+c^2) tan B is equal to

Show that the orthocentre of DeltaABC having vertices A(x_1, y_1), B(x_2, y_2), C(x_3, y_3) is : (x_1 tan A + x_2 tan B + x_3 tan C)/(tan A + tan B + tan C) , (y_1 tan A + y_2 tan B + y_3 tan C)/(tan A + tan B + tan C))

If in Delta ABC, tan.(A)/(2) = (5)/(6) and tan.( C)/(2)= (2)/(5) , then prove that a, b, and c are in A.P.

In Delta ABC if tan(A/2) tan(B/2) + tan(B/2) tan(C/2) = 2/3 then a+c

If A+B+C=pi , prove that : tan( A/2) tan (B/2) + tan (B/2 )tan (C/2)+ tan( C/2) tan (A/2) =1

In any Delta ABC , prove that : ((a-b)/c) = (tan (A/2) - tan (B/2))/(tan (A/2) + tan (B/2)

In triangle ABC, if |[1,1,1], [cot (A/2), cot(B/2), cot(C/2)], [tan(B/2)+tan(C/2), tan(C/2)+ tan(A/2), tan(A/2)+ tan(B/2)]| then the triangle must be (A) Equilateral (B) Isoceless (C) Right Angle (D) none of these

The area of a DeltaABC is b^(2)-(c-a)^(2) . Then ,tan B =

By using above basic addition/ subtraction formulae, prove that (i) tan (A+B)=(tan A+tan B)/(1-tan A tan B) , (ii) tan (A-B)=(tan A-tanB)/(1+tan A tan B) (iii) sin2theta=2sintheta costheta , (iv) cos2theta=cos^(2)theta=1-2sin^(2)theta=2cos^(2)theta-1 (v) tan 2theta=(2 tan theta)/(1-tan^(2) theta)

In DeltaABC,(a+b+c)(tan(A/2)+tan (B/2))=