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Question

# If sinA=nsin(A+2B), then tan(B)tan(A+B) is equal to,

A
2nn+1
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B
2nn1
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C
nn+1
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D
1nn+1
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Solution

## The correct option is D 1−nn+1We have,sinA=nsin(A+2B)sin(A+2B)=sinAnNow,sin(A+2B)+sinA=2sin(A+2B+A)2cos(A+2B−A)2sin(A+2B)+sinA=2sin(A+B)cosBsinAn+sinA=2sin(A+B)cosBsinA(1n+1)=2sin(A+B)cosB ............(1)sin(A+2B)−sinA=2sin(A+2B−A)2cos(A+2B+A)2sin(A+2B)−sinA=2cos(A+B)sinBsinAn−sinA=2cos(A+B)sinBsinA(1n−1)=2cos(A+B)sinB ............(2)On dividing (1) by (2), we get(1+n)(1−n)=tan(A+B)tanB(1−n)(1+n)=tan(B)tan(A+B)Hence, this is the answer.

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