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If $cosec A + \sec A = cosec B + \sec B$ then $\tan A \tan B$ is equal to

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tan⁡A+B2

cot⁡A+B2

cot⁡A−B2

tan⁡A−B2

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detailed solution

Correct option is B

sec⁡A−sec⁡B=cosec⁡B−cosec⁡A⇒ cos⁡B−cos⁡Acos⁡Acos⁡B=sin⁡A−sin⁡Bsin⁡Asin⁡B⇒ sin⁡Asin⁡Bcos⁡Acos⁡B=sin⁡A−sin⁡Bcos⁡B−cos⁡A⇒ tan⁡Atan⁡B=2cos⁡A+B2sin⁡A−B22sin⁡A+B2sin⁡A−B2=cot⁡A+B2

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If cosec⁡A+sec⁡A=cosec⁡B+sec⁡B then tan⁡Atan⁡B is equal to

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If $cosec A + \sec A = cosec B + \sec B$ then $\tan A \tan B$ is equal to