Question

sin∅/1-cos∅ = cos∅+cot∅

Answer 1

♦Correct question♦:

To Prove:

$LHS = \frac{ \sin(θ) }{1 - \cos(θ) } \: and \: RHS = \cosec(θ) + \cot(θ)$

Proof:

Taking LHS:

$LHS = \frac{ \sin(θ) }{1 - \cos(θ) }$

On rationalising the denominator, we get:

$= \frac{ \sin(θ) }{1 - \cos(θ) } \times \frac{1 + \cos(θ) }{ 1 + \cos(θ) }$

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

$= \frac{ \sin(θ)(1 + \cos(θ)) }{1 - \cos^{2} (θ) }$

[∵ sin²θ+cos²θ=1]

$= \frac{ \sin(θ) (1 + \cos(θ) )}{ \sin^{2} (θ) }$

$= \frac{1 + \cos(θ) }{ \sin(θ) }$

$= \frac{1}{ \sin(θ) } + \frac{ \cos(θ) }{ \sin(θ) }$

[∵ cosecθ=1/sinθ], and

[∵ cotθ=cosθ/sinθ]

As,

$LHS = cosec(θ) + \cot(θ) = RHS$

Hence, proved.