Question

prove that = cos theta/1-tan theta+sin theta/1-cot theta = sin theta + cos theta​

Answer 1

(Instead of θ, I use A)

Answer:

Step-by-step explanation:

Given :

To prove :

$\frac{cosA}{1-tanA} +\frac{sinA}{1-cotA} = sinA+cosA$

Solution :

We know that,

$tanA = \frac{sinA}{cosA} ,cotA=\frac{cosA}{sinA}$

a² - b² = (a + b) (a - b)

Proof :

LHS = $\frac{cosA}{1-tanA} +\frac{sinA}{1-cotA}$

⇒ $\frac{cosA}{1-\frac{sinA}{cosA} } +\frac{sinA}{1-\frac{cosA}{sinA} }$

⇒ $\frac{cosA}{\frac{cosA-sinA}{cosA} } +\frac{sinA}{\frac{sinA-cosA}{sinA} }$

⇒ $\frac{cos^2A}{-(sinA-cosA) } +\frac{sin^2A}{sinA-cosA }$

⇒ $-\frac{cos^2A}{sinA-cosA} +\frac{sin^2A}{sinA-cosA}$

⇒ $\frac{sin^2A-cos^2A}{sinA - cosA} = \frac{(sinA+cosA)(sinA-cosA)}{sinA-cosA} = sinA+cosA$=RHS

Hence, proved

Answer 2

Solution:-

To Proof:-

$⇒ \frac{ \cos(θ) }{1 - \tan(θ) } + \frac{ \sin(θ) }{1 - \cot(θ) } = \sin(θ) + \cos(θ)$

Proof:-

L.H.S.

$⇒ \frac{ \cos(θ) }{1 - \tan(θ) } + \frac{ \sin(θ) }{1 - \cot(θ) } \\ \\ ⇒ \frac{ \cos(θ) }{1 - \frac{ \sin(θ) }{ \cos(θ) } } + \frac{ \sin(θ) }{1 - \frac{ \cos(θ) }{ \sin(θ) } } \\ \\ ⇒ \: \frac{ \cos(θ) }{ \frac{ \cos(θ) - \sin(θ) }{ \cos(θ) } } + \frac{ \sin(θ) }{ \frac{ \sin(θ) - \cos(θ) }{ \sin(θ) } } \\ \\ ⇒ \frac{ { \cos(θ) }^{2} }{ \cos(θ) - \sin(θ) } + \frac{ { \sin(θ) }^{2} }{ \sin(θ ) - \cos(θ) } \\ \\ ⇒\frac{ { \cos(θ) }^{2} }{ \cos(θ) - \sin(θ) } - \frac{ { \sin(θ) }^{2} }{ \cos(θ) - \sin(θ) } \\ \\ ⇒ \frac{ { \cos(θ) }^{2} - { \sin(θ) }^{2} }{ \cos(θ) - \sin(θ) } \\ \\ ⇒ \: \frac{( \cos(θ) - \sin(θ) )( \cos(θ) + \sin(θ)) }{ \cos(θ) - \sin(θ) } \\ \\ ⇒ \: \cos(θ) + \sin(θ)$

⇒ L.H.S. = R.H.S.

Hence Proved!

Identity Used:-

• ( cos²θ - sin²θ) = [ ( cosθ + sinθ)(cosθ -