Question

secA + tanA = x
sinA = ?
(A) 1 - 2x/1+x^2
(B) x^2 - 1/z^2 + 1
(C) 1 + x^2/2(1 - x^2)
(D) x^2 + 1/x^2 - 1

Answer 1

=> secA + tanA = X

=> secA + √tan²A = X

=> secA + √sec²A-1 = X

=> √sec²A-1 = X - secA

=> sec²A - 1 = (X-secA)²

=> sec²A - 1 = X² + sec²A - 2XsecA

=> 2XsecA = X² + 1

=> secA = X²+1/2X

=> cosA = 2X/(1+X²) = Base/hypotenuse

_______________________±

» We know that :-

=> (hypotenuse)² = (perpendicular)² + (Base)²

=> (1+X²)² = (Perpendicular)² + (2X)²

=> (Perpendicular)² = 1+X^4+2X²-4X²

=> (Perpendicular)² = X^4+1-2X²

=> (Perpendicular)² = (1-X²)²

=> Perpendicular = 1 - X²

_________________________

» Thus ,

=> SinA = Perpendicular/hypotenuse

$= > SinA = \frac{1 - {x}^{2} }{1 + {x}^{2} }$
_____________________[ANSWER]

●Hence the option "A" is the correct answer.

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_-_-_-_✌☆$BE \: \: \: \: BRAINLY$☆✌_-_-_-_

Answer 2

Heya....hope this helps u.....@skb