Question

$\sqrt{1 + sin a} \div \sqrt{1 -sin a } = sec a + tan a$
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Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\bf \: Prove \: that \: \sf \: \sqrt{\dfrac{1 + sina}{1 - sina} } = seca \: + \: tana$

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

Consider LHS,

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

On multiply and divide by 1 + sina, we get

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

$\rm \: = \: \sqrt{\dfrac{ {(1 + sina)}^{2} }{1 - {sin}^{2}a } } \: \: \{ \because \: {x}^{2} - {y}^{2} = (x + y)(x - y) \}$

$\rm \: = \: \dfrac{1 + sina}{ \sqrt{ {cos}^{2}a } } \: \: \: \: \: \: \: \: \: \{ \because \: 1 - {sin}^{2}a = {cos}^{2}a \}$

$\rm \: = \: \dfrac{1 + sina}{cosa}$

$\rm \: = \: \dfrac{1}{cosa} + \dfrac{sina}{cosa}$

$\rm \: = \: seca \: + \: tana$

$\rm \: = \: RHS$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

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