Question

$\huge\color{pink}\mathfrak{Answer}$

$\sf{Prove\:That\:cosecθ-cotθ=\frac{sinθ} {1+cosθ}}$ ​

Answer 1

$\sf{To\: Prove} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \bold{{\:cosecθ-cotθ=\frac{sinθ} {1+cosθ}}}$

__________________________

$\bold{LHS =cosecθ-cotθ }$

and

$\bold{RHS =\frac{sinθ} {1+cosθ} }$

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now, we solve LHS

$\bold{LHS =cosecθ-cotθ }$

and we know that,

$\boxed{\bold{cosec \theta = \frac{1}{sin \theta} \: \: and \: \: cot \theta = \frac{cos \theta}{sin \theta} } }$

$\bold{ : \implies LHS = \frac{1}{sin \theta} - \frac{cos \theta}{sin \theta} } \\ \\ \bold{: \implies \:LHS = \frac{1 - cos \theta}{sin \theta} } \: \: \: \: \: \:$

now, multiple (1 + cos theta) by both numerator and denominator

$\bold{: \implies LHS = \frac{(1 - cos \theta) \times (1 + cos \theta)}{sin \theta(1 + cos \theta)} } \\ \\ \bold{: \implies LHS = \frac{1 - {cos}^{2} \theta }{sin \theta(1 + cos \theta)} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

and we know that

$\bold{1 - {cos}^{2} \theta = {sin}^{2} \theta}$

$\bold{: \implies LHS = \frac{ \cancel{sin}^{2} \theta }{ \cancel{sin \theta}(1 + cos \theta)}} \: \: \: \: \\ \\ \bold{: \implies LHS = \frac{sin \theta}{1 + cos \theta} =RHS }$

hence, proved

Answer 2

Let the L.H.S of the given equation

LHS=

cosecθ−cotθ

1

−

sinθ

1

=

cosecθ−cotθ

1

×

cosecθ+cotθ

cosecθ+cotθ

−

sinθ

1

=

cosec

2

θ−cot

2

θ

cosecθ+cotθ

−

sinθ

1

=

1

cosecθ+cotθ

−

sinθ

1

=

sinθ

sinθ(cosecθ+cotθ)−1

=

sinθ

1+cosθ−1

=

sinθ

1

+

sinθ

cosθ

−

sinθ

1

=

sinθ

1

+cotθ−cosecθ

=

sinθ

1

−(cosecθ−cotθ)

=

sinθ

1

−(cosecθ−cotθ)×

(cosecθ+cotθ)

(cosecθ+cotθ)

=

sinθ

1

−

(cosecθ+cotθ)

1