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Q.

If cosα+cosβ=m and sinα+sinβ=n, then sin(α+β)=

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a

m2-n2m2+n2

b

n2-m2m2+n2

c

2mnm2-n2

d

2mnm2+n2

answer is C.

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Detailed Solution

m=cosα+cosβ=2cosα+β2cosα-β2

n=sinα+sinβ=2sinα+β2cosα-β2

 nm=tanα+β2 
Now,

sinα+β=2tanα+β21+tan2α+β2=2nm1+n2m2=2mnm2+n2


 

 

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