prove that cos(45° + A)÷ cos(45º – A) = sec2A-tan2A​

Question

prove that
cos(45° + A)÷
cos(45º – A) =
sec2A-tan2A​

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Adeline 1 month 2021-08-12T14:21:39+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-12T14:22:53+00:00

    Answer:

    LHS =

    \frac{ \cos(45 + x) }{ \cos(45 – x) }

    cos(45−x)

    cos(45+x)

    = \frac{ \cos(x) \cos(45) – \sin(x) \sin(45) }{ \cos(x) \cos(45) + \sin(x) \sin(45)}=

    cos(x)cos(45)+sin(x)sin(45)

    cos(x)cos(45)−sin(x)sin(45)

    = \frac{ \cos(x) \frac{1}{ \sqrt{2} } – \cos(x) \frac{1}{ \sqrt{2} } }{\cos(x) \frac{1}{ \sqrt{2} } + \cos(x) \frac{1}{ \sqrt{2} } }=

    cos(x)

    2

    1

    +cos(x)

    2

    1

    cos(x)

    2

    1

    −cos(x)

    2

    1

    = \frac{ \cos(x) – \sin(x) }{ \cos(x) + \sin(x) }=

    cos(x)+sin(x)

    cos(x)−sin(x)

    Multiplying numerator and denominator by cos(x)-sin(x)

    = \frac{ {( \cos(x) – \sin(x) )}^{2} }{( \cos(x) – \sin(x) )( \cos(x) + \sin(x) )}=

    (cos(x)−sin(x))(cos(x)+sin(x))

    (cos(x)−sin(x))

    2

    = \frac{ \cos {}^{2} (x) + \sin {}^{2} (x) – 2 \sin(x) \cos(x) }{ \cos {}^{2} (x)- \sin {}^{2} (x) }=

    cos

    2

    (x)−sin

    2

    (x)

    cos

    2

    (x)+sin

    2

    (x)−2sin(x)cos(x)

    = \frac{1 – \sin(2x) }{ \cos(2x) }=

    cos(2x)

    1−sin(2x)

    = \frac{1}{ \cos(2x) } – \frac{ \sin(2x) }{ \cos(2x) }=

    cos(2x)

    1

    cos(2x)

    sin(2x)

    = \sec(2x) – \tan(2x)=sec(2x)−tan(2x)

    = RHS

    Hence Proved

    Step-by-step explanation:

    Pllzz like my answer ♥️

    0
    2021-08-12T14:23:00+00:00

    Step-by-step explanation:

    LHS =

     \frac{ \cos(45 + x) }{ \cos(45 - x) }

     =  \frac{ \cos(x)  \cos(45)  - \sin(x) \sin(45)   }{ \cos(x)  \cos(45) +  \sin(x) \sin(45)}

     =  \frac{ \cos(x) \frac{1}{ \sqrt{2} }   -  \cos(x)  \frac{1}{ \sqrt{2} } }{\cos(x) \frac{1}{ \sqrt{2} }    +   \cos(x) \frac{1}{ \sqrt{2} } }

     =  \frac{ \cos(x)  -   \sin(x) }{ \cos(x)  +   \sin(x) }

    Multiplying numerator and denominator by cos(x)-sin(x)

     =  \frac{ {( \cos(x)  -  \sin(x) )}^{2} }{( \cos(x)  - \sin(x)  )( \cos(x)  +  \sin(x) )}

     =  \frac{ \cos {}^{2} (x)  +  \sin {}^{2} (x) - 2 \sin(x)  \cos(x)  }{ \cos {}^{2}  (x)- \sin {}^{2} (x)  }

     =  \frac{1 -  \sin(2x) }{ \cos(2x) }

     =  \frac{1}{ \cos(2x) }  -  \frac{ \sin(2x) }{ \cos(2x) }

     =  \sec(2x)  -  \tan(2x)

    = RHS

    Hence Proved

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