Question

Hey !!

TRIGONOMETRIC RATIOS :

Prove that :

√(1+cos∅ / 1-cos∅) = cosec∅ + cot∅

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Answer 1

To Prove

√(1+cos∅ / 1-cos∅) = cosec∅ + cot∅

R.H.S → cosec∅ + cot∅

Since we know that,

• cosec∅ = 1/sin∅
• cot ∅ = cos∅/sin∅

→ 1/sin∅ + cos∅/sin∅

→ (1 + cos∅)/sin∅

→ (1 + cos∅)/sin∅ = √[(1 + cos∅)/(1 - cos∅)]

→ (1 + cos∅)²/sin²∅ = (1 + cos∅)/(1 - cos∅)

→ (1 + cos∅)(1 + cos∅)/(1 - cos∅)(1 + cos∅) = (1 + cos∅)/(1 - cos∅)

→ (1 + cos∅)/(1 - cos∅) = (1 + cos∅)/(1 - cos∅)

Remember

• sin²∅ = 1 - cos²∅
• 1² - cos²∅ = (1 - cos∅)(1 + cos∅)

Hence Proved

Answer 2

$\sqrt{ \dfrac{1 \: + \: cos \: \theta}{1 \: - \: cos \: \theta} }$ = cosec Ø + cot Ø

• We have to prove L.H.S. = R.H.S

» Take L.H.S.

=> $\sqrt{ \dfrac{1 \: + \: cos \: \theta}{1 \: - \: cos \: \theta} }$

Rationalize it.

=> $\sqrt{ \dfrac{1 \: + \: cos \: \theta}{1 \: - \: cos \: \theta} }$ × $\sqrt{ \dfrac{1 \: + \: cos \: \theta}{1 \: + \: cos \: \theta} }$

» (a + b) (a + b) = a² + b²

(a + b) (a - b) = a² - b²

=> $\sqrt{ \dfrac{ {(1 \: + \: cos \: \theta)}^{2} }{ {1}^{2} \: - \: {cos }^{2} \theta } }$

» sin²Ø + cos²Ø = 1

1 - cos²Ø = sin²Ø

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

=> $\dfrac{1\:+\:cos\:\theta}{sin\:\theta}$

=> $\dfrac{1}{sin\:\theta}$ + $\dfrac{cos\:\theta}{sin\:\theta}$

» cosec Ø = $\dfrac{1}{sin\:\theta}$

» cot Ø = $\dfrac{cos\:\theta}{sin\:\theta}$

=> cosec Ø + cot Ø

• L.H.S. = R.H.S.

_____________ [HENCE PROVED]