Question

can any one can solve it​

Answer 1

To Prove :-

• (cotA - cosA) / (cotA + cosA) = (cosecA - 1)/(cosecA + 1)

Formula used :-

• cotA = (cosA/sinA)
• (1/sinA) = cosecA

Solution :-

Solving LHS , we get,

→ (cotA - cosA) / (cotA + cosA)

Putting cotA = (cosA/sinA) in Numerator & Denominator we get,

→ [ (cosA/sinA) - cosA ] / [ (cosA/sinA) + cosA ]

Taking cosA common From Numerator & Denominator now,

→ cosA [ (1/sinA) - 1 ] / cosA [ (1/sinA) + 1 ]

→ [ (1/sinA) - 1 ] / [ (1/sinA) + 1 ]

Putting (1/sinA) = cosecA now, we get,

→ (cosecA - 1) / (cosecA + 1) = RHS (Hence Proved).

Answer 2

Solution

$\dfrac{Cot\: \theta\: - \: Cos\theta}{Cot\: \theta\: + \: Cos\: \theta}\: = \: \dfrac{Cosec\: \theta\: - \: 1}{Cosec\: \theta\: + \: 1}$

Formula

$Cot\: \theta \: = \: \dfrac{Cos\:\theta}{Sin\: \theta}$

Now put same value in equation no 1

Take L.H.S

$\rightarrow\: \dfrac{\dfrac{Cos\: \theta}{Sin\: \theta}\: - \: Cos\: \theta}{\dfrac{Cos\: \theta}{Sin\: \theta}\: +\: Cos\: \theta}$

$\rightarrow\:\dfrac{\dfrac{Cos\: \theta\: - \: Sin\: \theta.Cos\: \theta}{Sin\: \theta}}{\dfrac{Cos\: \theta\: + \: Sin \: \theta.Cos\: \theta}{Sin\: \theta}}$

$\rightarrow\:\dfrac{Cos\: \theta\: - \: Sin\: \theta.Cos\: \theta}{Cos\: \theta\: + \: Sin \: \theta.Cos\: \theta}$

$\rightarrow\:\dfrac{Cos\: \theta(1\: - \: Sin\: \theta)}{Cos\: \theta(1\:+ \: Sin \: \theta)}$

$\rightarrow\:\boxed{\dfrac{1\: - \: Sin\:\theta}{1\: + \: Sin\: \theta}}$

Take R.H.S

$\rightarrow\:\dfrac{Cosec\: \theta\: - \: 1}{Cosec\: \theta\: + \: 1}$

$\rightarrow\:\dfrac{\dfrac{1}{Sin\: \theta}\: - \: 1}{\dfrac{1}{Sin\: \theta}\: + \: 1}$

$\rightarrow\: \dfrac{\dfrac{1\: - \: Sin\: \theta}{Sin\: \theta}}{\dfrac{1\: + \: Sin\: \theta}{Sin\: \theta}}$

$\rightarrow\:\boxed{\dfrac{1\: - \: Sin\: \theta}{1\: + \: Sin\: \theta}}$

$\boxed{\boxed{L.H.S\: = \: R.H.S}}$

Hence Proved