Question

Prove that:
√(1+cosA) / √(1-cosA) = (cosecA + cotA)

Answer 1

Answer:

Question:

Prove that

$\bf \: \sqrt{\dfrac{1 + cosA}{1 - cosA} } = cosecA + cotA$

$\bf\underbrace\orange{Solution:}$

Consider LHS

$\bf \: \sqrt{\dfrac{1 + cosA}{1 - cosA} }$

On rationalize the denominator, we get

$\bf \: ⟹\sqrt{\dfrac{1 + cosA}{1 - cosA} \times \dfrac{1 + cosA}{1 + cosA} }$

$\bf \: ⟹\sqrt{\dfrac{(1 + {cosA)}^{2} }{1 - {cos}^{2}A } }$

$\bf \: ⟹\sqrt{\dfrac{(1 + {cosA)}^{2} }{{sin}^{2}A } }$

$\bf\implies \:\dfrac{1 + cosA}{sinA}$

$\bf\implies \:\dfrac{1}{sinA} + \dfrac{cosA}{sinA}$

$\bf\implies \:cosecA + cotA$

$\bf\large \underbrace\red{Hence, \: proved}$

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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

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