English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove (sec(x)-tan(x))(1+sin(x))=cos(x)

الحلّ

prove

(sec (x) - tan (x)) (1 + sin (x)) = cos (x)

صحيح

خطوات الحلّ

(sec (x) - tan (x)) (1 + sin (x)) = cos (x)

قم بالعمل على الطرف الأيسر

(sec (x) - tan (x)) (1 + sin (x))

sin, cos :

عبّر بواسطة

(1 + sin (x)) (sec (x) - tan (x))

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

:Use the basic trigonometric identity

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

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

:Use the basic trigonometric identity

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

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

بسّط:

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

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

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

بسّط:

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

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

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

فعّل القانون

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

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

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

:اضرب كسور

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

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

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

Rewrite using trig identities

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

(1 + sin (x)) (1 - sin (x))

وسٌع:

1 - sin^{2} (x)

(1 + sin (x)) (1 - sin (x))

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

فعّل قانون فرق المربّعات

a = 1, b = sin (x)

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

1^{a} = 1

فعّل القانون

1^{2} = 1

= 1 - sin^{2} (x)

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

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

:فعّل نطريّة فيتاغوروس

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

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

cos (x) :

إلغ العوامل المشتركة

= cos (x)

= cos (x)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e tan (- x) = cos (x)

p ro v e cos (4 5^{\circ} - x) = cos (x - 4 5^{\circ})

p ro v e \frac{1 - csc ^{2} ( A )}{csc ^{2} ( A )} = - cos^{2} (A)

p ro v e sin^{0.5} (x) cos (x) - sin^{2.5} (x) cos (x) = cos^{3} (x) sin (x)

p ro v e sin (\frac{2 π}{3}) = \frac{3}{2}