cos a – sin a +1 / cos a +.sin a -1 =cosec a +cot a. prove it​

Question

cos a – sin a +1 / cos a +.sin a -1 =cosec a +cot a. prove it​

in progress 0
Iris 1 month 2021-08-14T20:21:03+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-14T20:22:52+00:00

    TO PROVE :

      \\ \implies { \bold{ \dfrac{ \cos(a) -  \sin(a) + 1}{ \cos(a)  +  \sin(a)  - 1}  = cosec(a) +  \cot(a) }} \\

    SOLUTION :

    • Let’s take L.H.S. –

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cos(a) -  \sin(a) + 1}{ \cos(a)  +  \sin(a)  - 1}  }} \\

    • Rationalization of denominator –

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cos(a) -  \sin(a) + 1}{ \cos(a)  +  \sin(a)  - 1}  \times  \dfrac{ \cos(a) +  \sin(a)  + 1}{\cos(a) +  \sin(a)  + 1}  }} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \{ \cos(a) -  \sin(a) + 1 \} \{\cos(a) +  \sin(a)  + 1 \}}{  \{\cos(a)  +  \sin(a) \}^{2}   -  {(1)}^{2} }}} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cos^{2} (a)  +  \sin(a) \cos(a)  +  \cos(a)  -  \sin(a) \cos(a) -  \sin^{2} (a)  -  \sin(a) + \cos(a) +  \sin(a)  + 1}{ \cos^{2} (a)  +  \sin^{2} (a) + 2 \sin(a) \cos(a)  - 1}}} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cos^{2} (a) + 2 \cos(a)  -  \sin^{2} (a) + 1}{ 1 + 2 \sin(a) \cos(a)  - 1}}} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cos^{2} (a) + 2 \cos(a)  -  \sin^{2} (a) + 1}{2 \sin(a) \cos(a) }}} \\

    • We should write this as –

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cos^{2} (a) + 2 \cos(a)  - (1 -  \cos^{2} (a)) + 1}{2 \sin(a) \cos(a) }}} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cos^{2} (a) + 2 \cos(a)  - 1  +  \cos^{2} (a) + 1}{2 \sin(a) \cos(a) }}} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{2 \cos^{2} (a) + 2 \cos(a)}{2 \sin(a) \cos(a) }}} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cancel{2 \cos(a)} \{  \cos (a) + 1 \}}{ \cancel{2 \cos(a)}  \{  \sin(a) \}}}} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{ \cos (a) + 1}{ \sin(a) }}} \\

      \\  \:  \:  =  \:  \:  { \bold{ \dfrac{1}{ \sin(a) }  +  \dfrac{ \cos (a) }{ \sin(a) }}} \\

      \\  \:  \:  =  \:  \:  { \bold{cosec(a) +  \cot(a) }} \\

      \\  \:  \:  =  \:  \:  { \bold{R.H.S.}} \\

      \\  \:  \: \: \: \:  \:  { \underbrace{ \bold{Hence \:  \: proved}}} \\

    0
    2021-08-14T20:22:55+00:00

    Step-by-step explanation:

    {\red{\underline{\underline{\bold{To\:prove:-}}}}}

    \frac{ \cos(a) -  \sin(a)  + 1 }{ \cos(a) +  \sin(a) - 1 }= cosec(a) - cot(a)

    {\blue{\underline{\underline{\bold{Proof:-}}}}}

    Let us verify by simplifying L.H.S and R.H.S separately.

    L.H.S =   \frac{ \cos(a) -  \sin(a)  + 1 }{ \cos(a)  +  \sin(a)  - 1}

    Divide throughout by sin(a):

    = \frac{ \frac{ \cos(a) }{ \sin(a) }  -  \frac{ \sin(a) }{ \sin(a) } +  \frac{1}{ \sin(a) }  }{ \frac{ \cos(a) }{ \sin(a) } +  \frac{ \sin(a) }{ \sin(a) }  -  \frac{1}{ \sin( a) }  } \\\\ =  \frac{ \cot(a)  - 1 +  \cosec(a) }{ \cot(a) - 1 +  \cosec(a)  } \\\\ =  \frac{ [\cot( a) - (1 -  \cosec(a) )][(\cot( a) - (1 -  \cosec(a) )] }{ [\cot(a )  + (1 -  \cosec(a))][( \cot( a) - (1 -  \cosec(a) )]} \\\\= \frac{ {[ \cot(a) - 1 +   \cosec(a) ] }^{2} }{ { [\cot(a)] }^{2}  -  {[1 -  \cosec(a)] }^{2} }  \\\\ = \frac{ \cot(a) + 1 +  {cosec}^{2}(a)  - 2 \cot(a)   - 2 \cosec(a) + 2 \cot( a)  \cosec(a)   }{  {cot}^{2}a - (1  +  {cosec}^{2}  a + 2 \cosec(a) )}  \\\\ = \frac{2 {cosec}^{2}a + 2 \cot(a)   \cosec(a)   - 2 \cot(a) - 2 \cosec(a)   }{ {cot}^{2} a - 1 -  {cosec}^{2}a + 2 \cosec(a)  } \\\\  = \frac{2 \cosec(a)( \cosec(a) +  \cot(a)  )  - 2 \cot(a) +  \cosec(a)  }{ {cot}^{2} a -  {cosec}^{2}a  - 1 +2 \cosec(a)  }\:  \:  \:  \: ( \therefore {cot}^{2}a -  {cosec}^{2}  a \:  =  - 1) \\\\ =  \frac{ (\cosec(a) +  \cot(a))(2 \cosec(a) - 2 )  }{ - 1 - 1  + 2 cosec(a) } \\\\ =  \frac{ \cosec(a)  +  \cot(a) (2 cosec(a)  - 2)}{(2cosec(a) - 2)} \\\\ = cosec(a) + cot(a)

    = R.H.S

    Hence proved

Leave an answer

Browse
Browse

18:9+8+9*3-7:3-1*13 = ? ( )