Question

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$\bold{Prove \: that \: \frac{1 + sinθ}{1 - sinθ} = {(secθ + tan)}^{2}}$


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Answer 1

Prove that :-

$\rm :\longmapsto\:\dfrac{1 + sin\theta}{1 - sin\theta} = {(sec\theta + tan\theta)}^{2}$

Identities Used :-

$\red{ \boxed{ \sf \: {sin}^{2}\theta + {cos}^{2}\theta = 1}}$

$\red{ \boxed{ \sf \: \dfrac{1}{cos\theta} = sec\theta}}$

$\red{ \boxed{ \sf \: \dfrac{sin\theta}{cos\theta} = tan\theta}}$

Solution :-

Consider,

$\rm :\longmapsto\:\dfrac{1 + sin\theta}{1 - sin\theta}$

On rationalizing the denominator, we get

$\sf \: = \: \: \dfrac{1 + sin\theta}{1 - sin\theta} \times \dfrac{1 + sin\theta}{1 + sin\theta}$

$\sf \: = \: \: \dfrac{ {(1 + sin\theta)}^{2} }{ {1}^{2} - {sin}^{2}\theta}$

$\: \: \: \: \: \: \: \: \: \because \: \red{ \boxed{ \sf \: {x}^{2} - {y}^{2} = (x + y)(x - y)}}$

$\sf \: = \: \: \dfrac{ {(1 + sin\theta)}^{2} }{ {1} - {sin}^{2}\theta}$

$\sf \: = \: \: \dfrac{ {(1 + sin\theta)}^{2} }{{cos}^{2}\theta}$

$\sf \: = \: \: {\bigg(\dfrac{1 + sin\theta}{cos\theta} \bigg) }^{2}$

$\sf \: = \: \: {\bigg(\dfrac{1}{cos\theta} + \dfrac{sin\theta}{cos\theta} \bigg) }^{2}$

$\sf \: = \: \: {(sec\theta + tan\theta)}^{2}$

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

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

Answer 2

Answer:

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Step-by-step explanation:

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