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Question

If sinA=nsinB, then (n-1)(n+1)tan(A+B)2=


A

sinA-B2

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B

tanA-B2

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C

cotA-B2

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D

None of these

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Solution

The correct option is B

tanA-B2


Explanation for the correct option.

Given that, sinA=nsinB

It can be rewritten as: sinAsinB=n1

By componendo and dividendo rule we have:
(sinA+sinB)(sinA-sinB)=(n+1)(n-1)
Now we know that,sinx+siny=2sin(x+y)2cos(x-y)2andsinx-siny=2sin(x-y)2cos(x+y)2

So we have
2sin(A+B)2cos(A-B)22sin(A-B)2cos(A+B)2=(n+1)(n-1)tan(A+B)2tan(A-B)2=(n+1)n-1n-1n+1tanA+B2=tanA-B2
Hence, option B is correct.


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