sec(A+B)/(2)=cosec(C)/(2)

If I is the incenter of a triangle ABC, then the ratio IA:IB:IC is equal to cosec (A)/(2):csc(B)/(2):csc(C)/(2)sin(A)/(2):sin(B)/(2):sin(C)/(2)sec(A)/(2):sec(B)/(2):sec(C)/(2) none of these

If A, B, C are interior angles of DeltaABC , the prove that cosec((A+B)/2)=sec(C/2) .

The angle A of the triangle ABC is obtuse, if sec(B+C)=cosec (B-C)=2,find the angles.

(1-sec^(2)A)/(cosec^(2)A-1) equals:

Prove the following identities: (1+tan A tan B)^(2)+(tan A-tan B)^(2)=sec^(2)A sec^(2)B(tan A+csc B)^(2)-(cot B-sec A)^(2)=2tan A cot B(csc A+sec B)

If cotθ = sqrt6 , then the value of is: (cosec^(2)θ + sec^(2)θ)/(cosec^(2)θ − sec^(2)θ) यदि cotθ = sqrt6 है तो (cosec^(2)θ + sec^(2)θ)/(cosec^(2)θ − sec^(2)θ) का मान हैः

The expression cosec^(2)A cot^(2)A-sec^(2)A tan^(2)A-(cot^(2)A-tan^(2)A)(sec^(2)A cosec^(2)A-1) is equal to

The expression cosec^(2)A cot^(2)A-sec^(2)A tan^(2)A-(cot^(2)A-tan^(2)A)(sec^(2)A cosec^(2)A-1) is equal to

If 2sec^(2) A - sec^(4) A - 2cosec^(2) A + cosec^(4) A = 15//4 , then tan A is equal to