If secA+tanA=3, find sinA+secA.
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Question= tana+ seca=3 then find
sina
Solution:
Given tanA+secA=3
so,sinA/cosA +1/cosA =3
sinA+1=3cosA
squaring on both sides,
1+(sinA)^2 +2sinA =9(cosA)^2
1+(sinA)^2 +2sinA =9(1-sinA^2)
SO, 10sinA^2+2sinA-8=0
5sinA^2 +sinA-4=0
SOLVING quadratic equation in sinA, we get sinA=4/5 OR sinA=-1
BUT WHEN sinA=-1 both tanA AND secA tends to infinity and will not give summation as 3 , so ,sinA has to be equal to be 4/5.
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