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العربية

prove (csc(α)+cot(α))(sec(α)-1)=tan(α)

الحلّ

prove

(csc (α) + cot (α)) (sec (α) - 1) = tan (α)

صحيح

خطوات الحلّ

(csc (α) + cot (α)) (sec (α) - 1) = tan (α)

قم بالعمل على الطرف الأيسر

(csc (α) + cot (α)) (sec (α) - 1)

sin, cos :

عبّر بواسطة

(- 1 + sec (α)) (cot (α) + csc (α))

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

:Use the basic trigonometric identity

= (- 1 + \frac{1}{cos ( α )}) (cot (α) + csc (α))

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

:Use the basic trigonometric identity

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

csc (x) = \frac{1}{sin ( x )}

:Use the basic trigonometric identity

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

(- 1 + \frac{1}{cos ( α )}) (\frac{cos ( α )}{sin ( α )} + \frac{1}{sin ( α )})

بسّط:

\frac{( - cos ( α ) + 1 ) ( cos ( α ) + 1 )}{cos ( α ) sin ( α )}

(- 1 + \frac{1}{cos ( α )}) (\frac{cos ( α )}{sin ( α )} + \frac{1}{sin ( α )})

- 1 + \frac{1}{cos ( α )}

وحّد:

\frac{- cos ( α ) + 1}{cos ( α )}

- 1 + \frac{1}{cos ( α )}

1 = \frac{1 cos ( α )}{cos ( α )}

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

1 \cdot cos (α) = cos (α) :

اضرب

= \frac{- cos ( α ) + 1}{cos ( α )}

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

\frac{cos ( α )}{sin ( α )} + \frac{1}{sin ( α )}

وحّد الكسور:

\frac{cos ( α ) + 1}{sin ( α )}

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

فعّل القانون

= \frac{cos ( α ) + 1}{sin ( α )}

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

(a) = a

:احذف الأقواس

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

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

:اضرب كسور

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

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

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

Rewrite using trig identities

\frac{( 1 + cos ( α ) ) ( 1 - cos ( α ) )}{cos ( α ) sin ( α )}

(1 + cos (α)) (1 - cos (α))

وسٌع:

1 - cos^{2} (α)

(1 + cos (α)) (1 - cos (α))

(a + b) (a - b) = a^{2} - b^{2}

فعّل قانون فرق المربّعات

a = 1, b = cos (α)

= 1^{2} - cos^{2} (α)

1^{a} = 1

فعّل القانون

1^{2} = 1

= 1 - cos^{2} (α)

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

1 = cos^{2} (α) + sin^{2} (α)

:فعّل نطريّة فيتاغوروس

1 - cos^{2} (α) = sin^{2} (α)

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

sin (α) :

إلغ العوامل المشتركة

= \frac{sin ( α )}{cos ( α )}

= \frac{sin ( α )}{cos ( α )}

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

:Use the basic trigonometric identity

= tan (α)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e 1 - 2 sin^{2} (θ) = 2 cos^{2} (θ) - 1

p ro v e sin (\frac{π}{4}) = cos (\frac{π}{4})

p ro v e sin (3 a) + sin (a) = 4 sin (a) - 4 sin (3 a)

p ro v e tan (β) = \frac{cos ( β )}{sin ( β ) cot ^{2} ( β )}

p ro v e \frac{sin ^{2} ( x )}{cos ( x )} = sin (x) tan (x)