Question

Prove that: √(sec x + 1)/(sec x - 1) = 1/(cosec x - cot x) ​

Answer 1

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

Consider LHS

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

can be rewritten as

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

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

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

On rationalizing the numerator, we get

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

$\rm \: = \: \sqrt{\dfrac{1 - {cos}^{2} x}{ {(1 - cosx)}^{2} } }$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}}}$

So, using this, we get

$\rm \: = \: \dfrac{ \sqrt{ {sin}^{2} x} }{1 - cosx}$

$\rm \: = \: \dfrac{ sinx }{1 - cosx}$

$\rm \: = \: \dfrac{1}{ \: \: \: \: \dfrac{1 - cosx}{sinx} \: \: \: \: }$

$\rm \: = \: \dfrac{1}{ \: \: \: \: \dfrac{1}{sinx} - \dfrac{cosx}{sinx} \: \: \: \: }$

$\rm \: = \: \dfrac{1}{ \: \: \: \: cosecx - cotx \: \: \: \: }$

Hence, Proved

$\rm \implies\:\: \boxed{\tt{ \sqrt{\dfrac{secx + 1}{secx - 1} } = \frac{1}{cosecx - cotx}}}$

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MORE TO KNOW

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

Answer 2

Answer:

Sorry Bro

Step-by-step explanation:

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