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العربية

prove ((tan^2(x)))/(sec(x)+1)=((1-cos(x)))/(cos(x))

الحلّ

prove

\frac{( tan ^{2} ( x ) )}{sec ( x ) + 1} = \frac{( 1 - cos ( x ) )}{cos ( x )}

صحيح

خطوات الحلّ

\frac{tan ^{2} ( x )}{sec ( x ) + 1} = \frac{1 - cos ( x )}{cos ( x )}

قم بالعمل على الطرف الأيسر

\frac{tan ^{2} ( x )}{sec ( x ) + 1}

sin, cos :

عبّر بواسطة

\frac{tan ^{2} ( x )}{1 + sec ( x )}

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

:Use the basic trigonometric identity

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

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

:Use the basic trigonometric identity

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

\frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{1 + \frac{1}{c o s ( x )}}

بسّط:

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

\frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{1 + \frac{1}{c o s ( x )}}

1 + \frac{1}{cos ( x )}

وحّد:

\frac{cos ( x ) + 1}{cos ( x )}

1 + \frac{1}{cos ( x )}

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

:حوّل الأعداد لكسور

= \frac{1 \cdot cos ( x )}{cos ( x )} + \frac{1}{cos ( x )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{1 \cdot cos ( x ) + 1}{cos ( x )}

1 \cdot cos (x) = cos (x) :

اضرب

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

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

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

(\frac{sin ( x )}{cos ( x )})^{2}

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

:فعّل قانون القوى

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

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

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

:اقسم الكسور

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

cos (x) :

إلغ العوامل المشتركة

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

بسّط:

- \frac{cos ( x ) - 1}{cos ( x )}

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

1 - cos^{2} (x)

حلل إلى عوامل:

- (cos (x) + 1) (cos (x) - 1)

1 - cos^{2} (x)

- 1

قم باخراج العامل المشترك

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

cos^{2} (x) - 1

حلل إلى عوامل:

(cos (x) + 1) (cos (x) - 1)

cos^{2} (x) - 1

1^{2}

كـ

1

اكتب مجددًا

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

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

فعّل قانون فرق المربّعات

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

= (cos (x) + 1) (cos (x) - 1)

= - (cos (x) + 1) (cos (x) - 1)

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

cos (x) + 1 :

إلغ العوامل المشتركة

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

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

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

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

بسّط

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

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e \frac{1 + sin ( x )}{1 - sin ( x )} - \frac{1 - sin ( x )}{1 + sin ( x )} = \frac{4 tan ( x )}{cos ( x )}

p ro v e sin (3 x) = 4 sin (x) \cdot cos^{2} (x) - sin (x)

p ro v e sin (x - \frac{3 π}{2}) + sin (\frac{π}{2} + x) = 2 cos (x)

p ro v e \frac{1}{cos ( x ) \cdot sin ( x )} - tan (x) = cot (x)

p ro v e sec^{2} (x) cot^{2} (x) = csc^{2} (x)