Question

cos A-sin A/
cosA + Sin A
= cosec +cot
​

Answer 1

Step-by-step explanation:

cos(α)−sin(α)+1

=csc(α)+cot(α)

_______________

✨IDENTITY USED✨

\color{red} \boxed{\csc {}^{2} \alpha = > 1 + \cot {}^{2} \alpha }

csc

2

α=>1+cot

2

α

_______________

L.H.S

\frac{ \cos( \alpha ) - \sin( \alpha ) + 1 }{ \cos( \alpha ) + \sin( \alpha ) - 1 } = \csc( \alpha ) + \cot( \alpha )

cos(α)+sin(α)−1

cos(α)−sin(α)+1

=csc(α)+cot(α)

divide both numerator and denominator by sin A

\begin{gathered} \frac{ \frac{ \cos( \alpha ) }{ \sin( \alpha ) } - \frac{ \sin( \alpha ) }{ \sin( \alpha ) } + \frac{1}{ \sin( \alpha ) } }{ \frac{ \cos( \alpha ) }{ \sin( \alpha ) } + \frac{ \sin( \alpha ) }{ \sin( \alpha ) } - \frac{1}{ \sin( \alpha ) } } \\ \end{gathered}

sin(α)

cos(α)

+

sin(α)

sin(α)

−

sin(α)

1

sin(α)

cos(α)

−

sin(α)

sin(α)

+

sin(α)

1

\begin{gathered} = > > \frac{ \cot( \alpha ) - 1 + \csc( \alpha ) }{ \cot( \alpha ) + 1 - \csc( \alpha ) } \\ \\ = > > \frac{ (\cot( \alpha ) + \csc( \alpha )) - 1}{( \cot( \alpha ) + 1 - \csc( \alpha ) )} \\ \\ = > > \frac{ \cot( \alpha ) + \csc( \alpha ) - \csc ^{2} \alpha - \cot ^{2} \alpha }{ \cot( \alpha ) + 1 - \csc( \alpha ) } \end{gathered}

=>>

cot(α)+1−csc(α)

cot(α)−1+csc(α)

=>>

(cot(α)+1−csc(α))

(cot(α)+csc(α))−1

=>>

cot(α)+1−csc(α)

cot(α)+csc(α)−csc

2

α−cot

2

α

\boxed{1 = > \csc ^{2} \alpha - \cot ^{2} \alpha }

1=>csc

2

α−cot

2

α

Use (a²-b²)=>{a-b}{a+b}

\begin{gathered} a= > > \cot( \alpha ) \\ \\ b = > > \csc( \alpha ) \end{gathered}

a=>>cot(α)

b=>>csc(α)

\frac{ \cot( \alpha ) + \csc( \alpha ) - ( \csc( \alpha ) - \cot( \alpha ) \: )( \csc( \alpha ) + \cot( \alpha ) \: )}{ \cot( \alpha ) + 1 - \csc( \alpha ) }

cot(α)+1−csc(α)

cot(α)+csc(α)−(csc(α)−cot(α))(csc(α)+cot(α))

= > > \frac{ \cot( \alpha ) + \csc( \alpha )(1 - (\csc( \alpha ) - \cot( \alpha ) \: ) \: ) }{ \cot( \alpha ) + 1 - \csc( \alpha ) }=>>

cot(α)+1−csc(α)

cot(α)+csc(α)(1−(csc(α)−cot(α)))

= > > \frac{ \cot( \alpha ) + \csc( \alpha ) - ( \: \: 1 - \csc( \alpha ) + \cot( \alpha ) \: \: )}{ \cot( \alpha ) + 1 - \csc( \alpha ) }=>>

cot(α)+1−csc(α)

cot(α)+csc(α)−(1−csc(α)+cot(α))

= > > \frac{ \cot( \alpha ) + \csc( \alpha ) \: \: ( \: \: \cot( \alpha ) + 1 - \csc( \alpha ) \: \: )}{ \cot( \alpha ) + 1 - \csc( \alpha ) }=>>

cot(α)+1−csc(α)

cot(α)+csc(α)(cot(α)+1−csc(α))

= > > \color{blue} \boxed{\csc( \alpha ) + \cot( \alpha ) }=>>

csc(α)+cot(α)

Hence proved...