Question

Prove that √(cosec A + 1) / cosec A - 1) = sec A + tan A

Answer 1

Answer:

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

${ \longrightarrow{ \sf{ \sqrt{ \frac{CosecA+1}{CosecA - 1} }= SecA + TanA}}}$

From LHS,

${ \longrightarrow{ \sf{ \sqrt{ \frac{CosecA + 1}{CosecA - 1} } }}}$

By Rationalizing,

${ \longrightarrow{ \sf{ \sqrt{ \frac{CosecA+1}{CosecA -1} } ×\sqrt{ \frac{CosecA+1}{CosecA +1} }}}}$

${ \longrightarrow{ \sf{ \sqrt{ \frac{ {(CosecA+1)}^{2} }{ {(CosecA) }^{2} - {(1)}^{2} } } }}}$

${ \longrightarrow{ \sf{ \sqrt{ \frac{ {(CosecA+1)}^{2} }{ {Cosec}^{2}A - 1 } } }}}$

${ \longrightarrow{ \sf{ \sqrt{ \frac{ {(CosecA+1)}^{2} }{ {Cot}^{2} A} } }}}$

By Cancelling Squares and roots,

${ \longrightarrow{ \sf{ \frac{CosecA +1}{CotA} }}}$

By Expanding Fraction,

${ \longrightarrow{ \sf{ \frac{CosecA}{CotA} + \frac{1}{CotA} }}} \:$

${ \longrightarrow{ \sf{ \frac{CosecA}{CotA} +TanA}}}$

${ \longrightarrow{ \sf{ \frac{1}{SinA} × \frac{SinA}{CosA} +TanA}}}$

${ \longrightarrow{ \sf{ \frac{1}{CosA} +TanA}}}$

${ \longrightarrow{ \sf{SecA+ TanA}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: { \therefore{ \large{ \underline{ \mathcal{ \pmb{\rm{Hence \: Proved}}}}}}}$

Used Identities & Formula:-

• Cosec²θ - Cot²θ = 1

• 1/Cotθ = Tanθ

• 1/Sinθ = Cscθ

• 1/cosθ = Secθ

• a² - b² = (a + b) (a - b)

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Answer 2

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

Consider, LHS

$\rm :\longmapsto\: \sqrt{\dfrac{cosecA + 1}{cosecA - 1} }$

We know,

$\boxed{ \bf{ \: cosecx = \frac{1}{sinx}}}$

So, using this, we get

$\rm \: = \: \sqrt{\dfrac{\dfrac{1}{sinA} + 1 }{\dfrac{1}{sinA} - 1} }$

$\rm \: = \: \sqrt{\dfrac{1 + sinA}{1 - sinA} }$

On rationalizing the denominator, we get

$\rm \: = \: \sqrt{\dfrac{1 + sinA}{1 + sinA} \times \dfrac{1 + sinA}{1 + sinA} }$

We know,

$\boxed{ \bf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}$

$\rm \: = \: \sqrt{\dfrac{ {(1 + sinA)}^{2} }{ {1}^{2} - {sin}^{2}A } }$

$\rm \: = \: \sqrt{\dfrac{ {(1 + sinA)}^{2} }{1 - {sin}^{2}A } }$

We know,

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

So, using this, we get

$\rm \: = \: \dfrac{1 + sinA}{ \sqrt{ {cos}^{2} A} }$

$\rm \: = \: \dfrac{1 + sinA}{cosA}$

$\rm \: = \: \dfrac{1}{cosA} + \dfrac{sinA}{cosA}$

$\rm \: = \: secA \: + \: tanA$

Hence,

$\rm :\longmapsto\:\boxed{ \bf{ \: \sqrt{\dfrac{cosecA + 1}{cosecA - 1} } = secA + tanA}}$

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