If `f(theta)=|[sin^2A,cot A,1],[sin^2B,cosB,1],[sin^2C,cosC,1]|` , then (a) `t a n A+t a n B+c` (b)`cotAcotBcotC` (c)`sin^2A+sin^2B+sin^2C` (d)`0`

If f(theta)=|[sin^2A,cot A,1],[sin^2B,cotB,1],[sin^2C,cotC,1]| , then (a) t a n A+t a n B+t a n C (b) cotAcotBcotC (c) sin^2A+sin^2B+sin^2C (d) 0

If f(theta)=|sin^2A cot A1sin^2BcosB1sin^2CcosC1| , then t a n A+t a n B+c cotAcotBcotC sin^2A+sin^2B+sin^2C 0

If f(theta)=det[[sin^(2)A,cot A,1sin^(2)B,cos B,1sin^(2)B,cos B,1sin^(2)C,cos C,1]], then (a) sin^(2)A+sin B+c(b)cot A cot B cot C(c)sin^(2)A+sin^(2)B+sin^(2)C(d)0

If A+B+C = pi , prove that [[sin^2A,cotA,1],[sin^2B,cotB,1],[sin^2C,cotC,1]] =0

For any "Delta"A B C the value of determinant |sin^2\ \ A cot A1sin^2B cot B1sin^2\ \ C cot C1| is equal to- s in A s in B s in C b. 1 c. 0 d. s in A+s in B+s in C

For any "Delta"A B C the value of determinant |sin^2\ \ A cot A1sin^2B cot B1sin^2\ \ C cot C1| is equal to- s in A s in B s in C b. 1 c. 0 d. s in A+s in B+s in C

Prove that, |[sin A, cosA, sin(A+theta)],[sinB, cosB, sin(B+theta)],[sinC, cosc, sin(C+theta)]| =0

det [[1,1,1sin A, sin B, sin Csin ^ (2) A, sin ^ (2) B, sin ^ (2) C]] =

In any triangle ABC,sin^(2)A-sin^(2)B+sin^(2)C is always equal to (A) 2sin A sin B cos C(B)2sin A cos B sin C(C)2sin A cos B cos C(D)2sin A sin B sin C

In DeltaABC , prove that: a(cosB+cosC)=2(b+c)sin^(2)A/2 .