Question

prove that 1- cos thita/ sin thita= sin thita /1+cos thita​

Answer 1

To Prove:

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

Solution:

R.H.S

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

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

Answer 2

Answer:

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}

= sinθ1−cosθ

= sinθ(1+cosθ)(1−cosθ)(1+cosθ)

= sinθ(1+cosθ)1−cos 2 = sinθ(1+cosθ)sin 2 θ [∵sin 2 θ=1−cos 2 θ]= 1+cosθsinθ

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