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sin (theta) = 3sin(theta +2alpha) then the value of [tan (theta+ alpha) + 2tan (alpha)] = a).3 b). 2 c). -1 d). 0 e). 1

sin (theta) = 3sin(theta +2alpha) then the value of [tan (theta+ alpha) + 2tan (alpha)] =
a).3 b). 2 c). -1 d). 0 e). 1

Grade:11

1 Answers

Sankalp Das
25 Points
3 years ago
Let theta=a and alpha=bThen wehave to find .. [tan (a+ 2b) + 2tan (b)]......3sin(a+2b)=sina...sina/sin(2a+b) =3....Applying cand d..2sin(a+b)cosa/-2cos(a+b)sina=4÷2...-tan(a+b)÷tana=2...tan(a+b)=-2tana....Putting in to find.....2tana-2tana=0...D is correct

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