Question

sin theta upon 1 minus cos theta is equal to cos theta + cot theta prove​

Answer 1

Step-by-step explanation:

$\sin( \beta ) \div 1 - \cos( \beta ) = \cos( \beta ) + \cot( \beta )$

take LHS

multiple sin(beta) up and down

than

${ \sin( \beta ) }^{2} = 1 - { \cos( \beta ) }^{2}$

than

$1 - { \cos( \beta ) }^{2} = (1 + \cos( \beta ))(1 - \cos( \beta ) )$

than it becomes

$(1 + \cos( \beta ) ) \div \sin( \beta )$

which is cesec+cot

Answer 2

AnswEr :

• To Prove :

$\dfrac{ \sin( \theta) }{1- \cos( \theta)} = \csc( \theta) + \cot( \theta)$

• Proof :

$\longrightarrow \dfrac{ \sin( \theta) }{1- \cos( \theta)}$

$\longrightarrow \dfrac{ \sin( \theta) }{1- \cos( \theta)} \times 1$

⠀⠀⠀⠀⋆ Multiplying by 1 + cos(θ) / 1+ cos(θ)

$\longrightarrow \dfrac{ \sin( \theta) }{1- \cos( \theta)} \times \dfrac{ 1 + \cos( \theta) }{1 + \cos( \theta)}$

$\longrightarrow \dfrac{ \sin( \theta) (1 + \cos( \theta))}{1- \cos^{2} ( \theta)}$

⠀⠀⠀⠀⋆ 1 - cos²(θ) = sin²(θ)

$\longrightarrow \dfrac{ \cancel{\sin( \theta)} (1 + \cos( \theta))}{ \cancel{\sin^{2} ( \theta)}}$

$\longrightarrow \dfrac{(1 + \cos( \theta))}{ \sin( \theta)}$

$\longrightarrow \dfrac{ 1}{ \sin( \theta)} + \dfrac{ \cos( \theta)}{ \sin( \theta)}$

⠀⠀⠀⠀⋆ 1 / sin(θ) = cosec(θ)

⠀⠀⠀⠀⋆ cos(θ) / sin(θ) = cot(θ)

$\longrightarrow \large\csc( \theta) + \cot( \theta)$

⠀

$\therefore \boxed{ \dfrac{ \sin( \theta) }{1- \cos( \theta)} = \csc( \theta) + \cot( \theta) }$