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شائع علم المثلثات >

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

الحلّ

prove

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

الحلّ

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيسر

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

sin, cos :

عبّر بواسطة

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

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

:Use the basic trigonometric identity

= \frac{sin ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{- 1 + \frac{1}{c o s ( θ )}}

\frac{sin ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{- 1 + \frac{1}{c o s ( θ )}}

بسّط:

\frac{sin ^{2} ( θ )}{- cos ( θ ) + 1}

\frac{sin ( θ ) \frac{s i n ( θ )}{c o s ( θ )}}{- 1 + \frac{1}{c o s ( θ )}}

- 1 + \frac{1}{cos ( θ )}

وحّد:

\frac{- cos ( θ ) + 1}{cos ( θ )}

- 1 + \frac{1}{cos ( θ )}

1 = \frac{1 cos ( θ )}{cos ( θ )}

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

1 \cdot cos (θ) = cos (θ) :

اضرب

= \frac{- cos ( θ ) + 1}{cos ( θ )}

= \frac{\frac{s i n ( θ )}{c o s ( θ )} sin ( θ )}{\frac{- c o s ( θ ) + 1}{c o s ( θ )}}

sin (θ) \frac{sin ( θ )}{cos ( θ )}

اضرب بـ:

\frac{sin ^{2} ( θ )}{cos ( θ )}

sin (θ) \frac{sin ( θ )}{cos ( θ )}

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

:اضرب كسور

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

a^{b} \cdot a^{c} = a^{b + c}

:فعّل قانون القوى

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

1 + 1 = 2 :

اجمع الأعداد

= sin^{2} (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{\frac{s i n ^{2} ( θ )}{c o s ( θ )}}{\frac{- c o s ( θ ) + 1}{c o s ( θ )}}

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

:اقسم الكسور

= \frac{sin ^{2} ( θ ) cos ( θ )}{cos ( θ ) ( - cos ( θ ) + 1 )}

cos (θ) :

إلغ العوامل المشتركة

= \frac{sin ^{2} ( θ )}{- cos ( θ ) + 1}

= \frac{sin ^{2} ( θ )}{- cos ( θ ) + 1}

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

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{1 - cos ( θ )}

cos^{2} (x) + sin^{2} (x) = 1

:فعّل نطريّة فيتاغوروس

sin^{2} (x) = 1 - cos^{2} (x)

= \frac{1 - cos ^{2} ( θ )}{1 - cos ( θ )}

\frac{1 - cos ^{2} ( θ )}{1 - cos ( θ )}

بسّط:

cos (θ) + 1

\frac{1 - cos ^{2} ( θ )}{1 - cos ( θ )}

1 - cos^{2} (θ)

حلل إلى عوامل:

- (cos (θ) + 1) (cos (θ) - 1)

1 - cos^{2} (θ)

- 1

قم باخراج العامل المشترك

= - (cos^{2} (θ) - 1)

cos^{2} (θ) - 1

حلل إلى عوامل:

(cos (θ) + 1) (cos (θ) - 1)

cos^{2} (θ) - 1

1^{2}

كـ

1

اكتب مجددًا

= cos^{2} (θ) - 1^{2}

x^{2} - y^{2} = (x + y) (x - y)

فعّل قانون فرق المربّعات

cos^{2} (θ) - 1^{2} = (cos (θ) + 1) (cos (θ) - 1)

= (cos (θ) + 1) (cos (θ) - 1)

= - (cos (θ) + 1) (cos (θ) - 1)

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

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

اختزل:

cos (θ) + 1

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

- cos (θ) + 1 = - (cos (θ) - 1)

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

بسّط

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

cos (θ) - 1 :

إلغ العوامل المشتركة

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= cos (θ) + 1

= 1 + cos (θ)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

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