[tex]\large\underline{\sf{Solution-}}[/tex]
Consider LHS
[tex]\rm :\longmapsto\: \sqrt{\dfrac{secx + 1}{secx - 1} } [/tex]
can be rewritten as
[tex]\rm \: = \: \sqrt{\dfrac{\dfrac{1}{cosx} + 1 }{\dfrac{1}{cosx} - 1} } [/tex]
[tex]\rm \: = \: \sqrt{\dfrac{\dfrac{1 + cosx}{cosx}}{\dfrac{1 - cosx}{cosx}} } [/tex]
[tex]\rm \: = \: \sqrt{\dfrac{1 + cosx}{1 - cosx} } [/tex]
On rationalizing the numerator, we get
[tex]\rm \: = \: \sqrt{\dfrac{1 + cosx}{1 - cosx} \times \dfrac{1 - cosx}{1 - cosx} } [/tex]
[tex]\rm \: = \: \sqrt{\dfrac{1 - {cos}^{2} x}{ {(1 - cosx)}^{2} } } [/tex]
We know,
[tex]\red{\rm :\longmapsto\:\boxed{\tt{ {sin}^{2}x + {cos}^{2}x = 1}}}[/tex]
So, using this, we get
[tex]\rm \: = \: \dfrac{ \sqrt{ {sin}^{2} x} }{1 - cosx} [/tex]
[tex]\rm \: = \: \dfrac{ sinx }{1 - cosx} [/tex]
[tex]\rm \: = \: \dfrac{1}{ \: \: \: \: \dfrac{1 - cosx}{sinx} \: \: \: \: } [/tex]
[tex]\rm \: = \: \dfrac{1}{ \: \: \: \: \dfrac{1}{sinx} - \dfrac{cosx}{sinx} \: \: \: \: } [/tex]
[tex]\rm \: = \: \dfrac{1}{ \: \: \: \: cosecx - cotx \: \: \: \: } [/tex]
Hence, Proved
[tex]\rm \implies\:\: \boxed{\tt{ \sqrt{\dfrac{secx + 1}{secx - 1} } = \frac{1}{cosecx - cotx}}}[/tex]
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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
