Math, asked by mahajabeenmomo, 3 months ago

if secA+tanA=x then what is the value of cosecA?

Answers

Answered by mathdude500
5

\large\underline{\sf{Given \:Question - }}

  \:  \:  \:  \:\sf \: secA + tanA = x

\large\underline{\sf{To\:Find - }}

 \sf \:   \: cosecA

\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}}  \end{gathered}

1. \:  \sf \:  {sec}^{2} A -  {tan}^{2} A = 1

2. \:  \:  \sf \: tanA =  \: \dfrac{sinA}{cosA}

3. \: \:   \sf \:  secA = \dfrac{1}{cosA}

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

Given that

\rm :\longmapsto\:secA + tanA = x -  -  - (1)

We know,

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

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

\rm :\longmapsto\:x \times (secA - tanA) = 1

\bf\implies \:secA - tanA = \dfrac{1}{x} -  -  - (2)

Now, Adding equation (1) and (2), we get

\rm :\longmapsto\:2secA = x + \dfrac{1}{x}

\bf\implies \:secA = \dfrac{ {x}^{2} + 1 }{2x} -  -  - (3)

Now, Subtracting equation (2) from equation (1), we get

\rm :\longmapsto\:2tanA = x - \dfrac{1}{x}

\bf\implies \:tanA = \dfrac{ {x}^{2} - 1 }{2x} -  -  - (4)

Now,

Consider,

\rm :\longmapsto\:sinA

 \sf \:  =  \: sinA \times \dfrac{cosA}{cosA}

 \sf \:  =  \: \dfrac{sinA}{cosA} \times cosA

 \sf \:  =  \: tanA \times \dfrac{1}{secA}

 \sf \:  =  \: \dfrac{ {x}^{2} - 1 }{\cancel{2x}} \times \dfrac{\cancel{2x}}{ {x}^{2} + 1 }

 \sf \:  =  \: \dfrac{ {x}^{2}  - 1}{ {x}^{2}  + 1}

Therefore,

\rm :\longmapsto\:cosecA = \dfrac{1}{sinA} = \dfrac{ {x}^{2}  + 1}{ {x}^{2}  - 1}

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

Similar questions