Question

$\frac{cos \: thita}{1 - sin \: thita} = \frac{1 + sin \: thita}{cos \: thita}$
​

Answer 1

Step-by-step explanation:

To Prove:

\sf \dfrac{1 - \cos \theta }{sin \: \theta} = \dfrac{ \sin \theta}{ 1+ \cos \theta}

sinθ

1−cosθ

=

1+cosθ

sinθ

Solution:

R.H.S

\begin{gathered} = \sf \dfrac{1 - \cos \theta }{sin \: \theta} \\ \\ = \frac{(1 - \cos \theta)(1 + \cos \theta)}{ \sin \theta(1 + \cos \theta)} \\ \\ = \frac{1 - \cos {}^{2} \theta}{ \sin \theta(1 + \cos \theta) } \\ \\ = \sf \frac{sin {}^{2} \theta }{ \sin \theta(1 + \cos \theta) } [ \: \: \because \: { \sin }^{2}\theta= 1 - { \cos }^{2} \theta] \\ \\ \sf = \frac{sin \theta}{1 \: + \cos\theta} \end{gathered}

=

sinθ

1−cosθ

=

sinθ(1+cosθ)

(1−cosθ)(1+cosθ)

=

sinθ(1+cosθ)

1−cos

2

θ

=

sinθ(1+cosθ)

sin

2

θ

[∵sin

2

θ=1−cos

2

θ]

=

1+cosθ

sinθ

Therefore, R.H.S = L.H.S [Proved]

Answer 2

$\Huge \color{magenta} {\mathbb {\underline {ANSWER:}}}$

$\normalsize \color{violet} {\tt {\frac { \cos(\theta)}{1- \sin(\theta)} = \frac{1+ \sin(\theta)}{\cos (theta)} ~is ~a ~true~statement.}}$

$\small \purple {\tt {The~ explanation~ has ~been ~attached~ above.}}$