Question

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

Answer 1

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

Answer 2

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:

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