If sin α=k sin β , then tan ((α-β)/2 ) cot ((α+β)/2 ) is equal to

Question

MARKS Content Team · Accepted Answer

Correct answer: (B) (k-1)/(k+1). Given sin α = k sin β , we have (sin α)/(sin β) = k . Applying componendo and dividendo: (sin α + sin β)/(sin α - sin β) = (k+1)/(k-1) Using the sum-to-product formulas sin A + sin B = 2 sin ((A+B)/2 ) cos ((A-B)/2 ) and sin A - sin B = 2 cos ((A+B)/2 ) sin ((A-B)/2 ) , we get: (2 sin ((α+β)/2 ) cos ((α-β)/2 ))/(2 cos ((α+β)/2 ) sin ((α-β)/2 )) = (k+1)/(k-1) This simplifies to tan ((α+β)/2 ) cot ((α-β)/2 ) = (k+1)/(k-1) . Taking the reciprocal of both sides: cot ((α+β)/2 ) tan ((α-β)/2 ) = (k-1)/(k+1) . Answer: (k-1)/(k+1)

If $\sin \alpha=k \sin \beta$, then $\tan \left(\dfrac{\alpha-\beta}{2}\right) \cot \left(\dfrac{\alpha+\beta}{2}\right)$ is equal to

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