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If (sinA)/(sinB)=(sqrt(3))/2"and"(cosA)/...

If `(sinA)/(sinB)=(sqrt(3))/2"and"(cosA)/(cosB)=(sqrt(5))/2` then `tanA+tanB` is equal to

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`sinA=sqrt(3)/2sinB`....(1) and `cosA=sqrt(5)/2cosB`....(2)
dividing (1) by(2) we get
`tanA=sqrt(3/5)tanB`
also squaring (1) and (2) and adding we get =>`3sin^(2)B+5cos^(2)B=4` as `sin^(2)A+cos^(2)A=1`...(3)
also substituting value of `5cos^(2)B` with help of eqn (3) we get=>`2sin^(2)B=1` thus `sinB=1/sqrt(2)` similarly `cosB=1/sqrt(2)` thus tanB=1,B=`pi/4` as for given range of angles their output will always be positve.
Thus `tanA=sqrt(3/5)` and `tanA+tanB=sqrt(3/5)+1`
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