secA+tanA=3/2 then sin A
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Hi !
Given,
SecA + tanA = 3/2
1/cosA + sinA/cosA = 3/2
sinA+1/cosA = 3/2
Squaring both sides,we get ,
(sinA +1)²/cos²A = 9/4
[cos²A can be written as 1 - sin²A]
(sinA +1) (sinA +1) / 1 - sin²A = 9/4
the denominator is of the form (a+b)(a-b) = a² - b²
(sinA +1) (sinA +1) / (1 - sinA)(1+ sinA) = 9/4
sinA + 1 / 1 - sinA = 9/4
cross multiplication,
4(sinA + 1) = 9( 1 - sinA)
4sinA + 4 = 9 - 9 sinA
13 sinA = 5
sinA = 5/13
Given,
SecA + tanA = 3/2
1/cosA + sinA/cosA = 3/2
sinA+1/cosA = 3/2
Squaring both sides,we get ,
(sinA +1)²/cos²A = 9/4
[cos²A can be written as 1 - sin²A]
(sinA +1) (sinA +1) / 1 - sin²A = 9/4
the denominator is of the form (a+b)(a-b) = a² - b²
(sinA +1) (sinA +1) / (1 - sinA)(1+ sinA) = 9/4
sinA + 1 / 1 - sinA = 9/4
cross multiplication,
4(sinA + 1) = 9( 1 - sinA)
4sinA + 4 = 9 - 9 sinA
13 sinA = 5
sinA = 5/13
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