Question

sin A / 1 + cos A
= cosec A - cot A
​

Answer 1

TO PROVE :–

$\\ \implies \bf \dfrac{\sin(A)}{1+ \cos(A)}=cosec(A)-\cot(A)$

SOLUTION :–

• Let's take L.H.S. –

$\\ \: \: = \: \: \bf \dfrac{\sin(A)}{1+ \cos(A)}$

• Now rationalization –

$\\ \: \: = \: \: \bf \dfrac{\sin(A)}{1+ \cos(A)} \times \dfrac{1 - \cos(A)}{1 - \cos(A)}$

$\\ \: \: = \: \: \bf \dfrac{ \{\sin(A) \} \{1 - \cos(A)\}}{\{1+ \cos(A) \} \{1 - \cos(A) \}}$

$\\ \: \: = \: \: \bf \dfrac{ \{\sin(A) \} \{1 - \cos(A)\}}{1 - \cos^{2} (A)}$

$\\ \: \: = \: \: \bf \dfrac{ \{\sin(A) \} \{1 - \cos(A)\}}{\sin^{2} (A)}$

$\\ \: \: = \: \: \bf \dfrac{1 - \cos(A)}{\sin(A)}$

$\\ \: \: = \: \: \bf \dfrac{1}{\sin(A)} - \dfrac{\cos(A)}{\sin(A)}$

$\\ \: \: = \: \: \bf cosec(A)-\cot(A)$

$\\ \: \: = \: \: \bf R.H.S.$

$\\ \implies \bf Hence \: \:Proved$

USED IDENTITY :–

$\\ \longrightarrow \bf 1 - { \cos}^{2}( \theta) = { \sin}^{2}( \theta)$

$\\ \longrightarrow \bf \dfrac{1}{\sin( \theta)} = cosec(\theta)$

$\\ \longrightarrow \bf \dfrac{\cos( \theta)}{\sin( \theta)} = \cot(\theta)$

Answer 2

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• We have to prove $\rightarrow \sf\frac{sin(A)}{1 + cos(A)} = cosec(A) - cot(A)$

${ \huge { \underline {\rm{Solution}}}}$

$\large \sf L.H.S. = \frac{sin(A)}{1+cos(A)}$

•Rationalise the denominator:

$\large \sf\rightarrow\frac{sin(A)}{1+cos(A)}×\frac{1-cos(A)}{1-cos(A)}$

$\large \sf \rightarrow \: \frac{ \{ \sin(A) \}}{ \{ + \cos(A) \}} \frac{ \{1 - \cos(A) \} }{ \{1 - cos(A) \}}$

$\large \sf \rightarrow \frac{ \{ \sin(A) \} \{1 - \cos(A) }{ \{ 1 - {cos}^{2} (A)\}}$

$\large \sf \rightarrow \frac{ \{ \sin(A \{ 1 - \cos(A) \}) \}}{ { \sin(A) }^{2} }$

$\sf \rightarrow \frac{1 - \cos(A) }{ \sin(A) } \\ \\ \sf \implies \frac{1}{ \sin(A) } - \frac{ \cos(A) }{ \sin(A) }$

$\implies \large \sf cosec(A) - \cot(A)$

$\sf\rightarrow \:R.H.S.$

#Hence proved

• Identity:

$\rightarrow \sf1 - { \cos}^{2} (θ) = { \sin }^{2} (θ)$

$\large \rightarrow \sf \frac{1}{ \sin(θ) } = cosec(θ)$

$\implies \sf \frac{cos(θ)}{ \sin(θ) } = \cot(θ)$

Hope it helps you...

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