lf A, B, C are angles of a triangle, then $$2sin$$⁡ $$A2cosec$$⁡ $$B2sin$$⁡ $$C2$$−$$sin$$⁡ Acot⁡ $$B2$$−$$cos$$⁡ A is

Correct option is 1independent of A, B, C

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lf A, B, C are angles of a triangle, then $2 \sin \frac{A}{2} cosec \frac{B}{2}$ $\sin \frac{C}{2} - \sin Acot \frac{B}{2} - \cos A$ is

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detailed solution

Correct option is A

2sinA2cosecB2sinC2−cosA2cosB2−cosA=2sinA2cosecB2×cosA+B2−cosA2cosB2−cosA=2sin⁡A2cosec⁡B2−sin⁡A2sin⁡B2−cos⁡A=−2sin2⁡A2−cos⁡A=−1

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lf A, B, C are angles of a triangle, then 2sin⁡ A2cosec⁡ B2sin⁡ C2−sin⁡ Acot⁡ B2−cos⁡ A is

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lf A, B, C are angles of a triangle, then $2 \sin \frac{A}{2} cosec \frac{B}{2}$ $\sin \frac{C}{2} - \sin Acot \frac{B}{2} - \cos A$ is