Question

if tan A + sec a is equal to X by Y then prove that cos A is equal to 2 x y divided by X square + Y square.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: secA + tanA = \dfrac{x}{y} - - - (1)$

We know,

$\rm \: {sec}^{2}A - {tan}^{2}A = 1$

can be rewritten as

$\rm \: (secA + tanA)(secA - tanA) = 1$

On substituting the value from equation (1), we get

$\rm \: \dfrac{x}{y}(secA - tanA) = 1$

$\rm\implies \:secA - tanA = \dfrac{y}{x} - - - (2)$

On adding equation (1) and (2), we get

$\rm \: 2secA = \dfrac{x}{y} + \dfrac{y}{x}$

$\rm \: 2secA = \dfrac{ {x}^{2} + {y}^{2} }{xy}$

$\rm \: secA = \dfrac{ {x}^{2} + {y}^{2} }{2xy}$

$\rm\implies \:cosA = \dfrac{2xy}{ {x}^{2} + {y}^{2} }$

Hence, Proved

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1