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prove (sec(θ)-1)/(sec(θ)+1)=(tan(θ)-sin(θ))/(tan(θ)+sin(θ))

الحلّ

prove

\frac{sec ( θ ) - 1}{sec ( θ ) + 1} = \frac{tan ( θ ) - sin ( θ )}{tan ( θ ) + sin ( θ )}

الحلّ

صحيح

خطوات الحلّ

\frac{sec ( θ ) - 1}{sec ( θ ) + 1} = \frac{tan ( θ ) - sin ( θ )}{tan ( θ ) + sin ( θ )}

قم بالعمل على الطرف الأيمن

\frac{tan ( θ ) - sin ( θ )}{tan ( θ ) + sin ( θ )}

sin, cos :

عبّر بواسطة

\frac{- sin ( θ ) + tan ( θ )}{sin ( θ ) + tan ( θ )}

tan (x) = \frac{sin ( x )}{cos ( x )}

:Use the basic trigonometric identity

= \frac{- sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}

\frac{- sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}

بسّط:

\frac{- cos ( θ ) + 1}{cos ( θ ) + 1}

\frac{- sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

وحّد:

\frac{sin ( θ ) cos ( θ ) + sin ( θ )}{cos ( θ )}

sin (θ) + \frac{sin ( θ )}{cos ( θ )}

sin (θ) = \frac{sin ( θ ) cos ( θ )}{cos ( θ )}

:حوّل الأعداد لكسور

= \frac{sin ( θ ) cos ( θ )}{cos ( θ )} + \frac{sin ( θ )}{cos ( θ )}

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{sin ( θ ) cos ( θ ) + sin ( θ )}{cos ( θ )}

= \frac{- sin ( θ ) + \frac{s i n ( θ )}{c o s ( θ )}}{\frac{s i n ( θ ) c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}

- sin (θ) + \frac{sin ( θ )}{cos ( θ )}

وحّد:

\frac{- sin ( θ ) cos ( θ ) + sin ( θ )}{cos ( θ )}

- sin (θ) + \frac{sin ( θ )}{cos ( θ )}

sin (θ) = \frac{sin ( θ ) cos ( θ )}{cos ( θ )}

:حوّل الأعداد لكسور

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

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

:بما أنّ المقامات متساوية، اجمع البسوط

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

= \frac{\frac{- s i n ( θ ) c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}{\frac{s i n ( θ ) c o s ( θ ) + s i n ( θ )}{c o s ( θ )}}

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

:اقسم الكسور

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

cos (θ) :

إلغ العوامل المشتركة

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

sin (θ)

قم باخراج العامل المشترك

= \frac{sin ( θ ) ( - cos ( θ ) + 1 )}{sin ( θ ) cos ( θ ) + sin ( θ )}

sin (θ)

قم باخراج العامل المشترك

= \frac{sin ( θ ) ( - cos ( θ ) + 1 )}{sin ( θ ) ( cos ( θ ) + 1 )}

sin (θ) :

إلغ العوامل المشتركة

= \frac{- cos ( θ ) + 1}{cos ( θ ) + 1}

= \frac{- cos ( θ ) + 1}{cos ( θ ) + 1}

= \frac{1 - cos ( θ )}{1 + cos ( θ )}

Rewrite using trig identities

cos (x) = \frac{1}{sec ( x )}

:Use the basic trigonometric identity

\frac{1 - \frac{1}{s e c ( θ )}}{1 + \frac{1}{s e c ( θ )}}

بسّط

\frac{1 - \frac{1}{s e c ( θ )}}{1 + \frac{1}{s e c ( θ )}}

1 + \frac{1}{sec ( θ )}

وحّد:

\frac{sec ( θ ) + 1}{sec ( θ )}

1 + \frac{1}{sec ( θ )}

1 = \frac{1 sec ( θ )}{sec ( θ )}

:حوّل الأعداد لكسور

= \frac{1 \cdot sec ( θ )}{sec ( θ )} + \frac{1}{sec ( θ )}

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{1 \cdot sec ( θ ) + 1}{sec ( θ )}

1 \cdot sec (θ) = sec (θ) :

اضرب

= \frac{sec ( θ ) + 1}{sec ( θ )}

= \frac{1 - \frac{1}{s e c ( θ )}}{\frac{s e c ( θ ) + 1}{s e c ( θ )}}

1 - \frac{1}{sec ( θ )}

وحّد:

\frac{sec ( θ ) - 1}{sec ( θ )}

1 - \frac{1}{sec ( θ )}

1 = \frac{1 sec ( θ )}{sec ( θ )}

:حوّل الأعداد لكسور

= \frac{1 \cdot sec ( θ )}{sec ( θ )} - \frac{1}{sec ( θ )}

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{1 \cdot sec ( θ ) - 1}{sec ( θ )}

1 \cdot sec (θ) = sec (θ) :

اضرب

= \frac{sec ( θ ) - 1}{sec ( θ )}

= \frac{\frac{s e c ( θ ) - 1}{s e c ( θ )}}{\frac{s e c ( θ ) + 1}{s e c ( θ )}}

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

:اقسم الكسور

= \frac{( sec ( θ ) - 1 ) sec ( θ )}{sec ( θ ) ( sec ( θ ) + 1 )}

sec (θ) :

إلغ العوامل المشتركة

= \frac{sec ( θ ) - 1}{sec ( θ ) + 1}

\frac{sec ( θ ) - 1}{sec ( θ ) + 1}

\frac{sec ( θ ) - 1}{sec ( θ ) + 1}

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

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