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شائع علم المثلثات >

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

الحلّ

prove

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

الحلّ

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيمن

(sec (x) - tan (x)) sec (x)

sin, cos :

عبّر بواسطة

(sec (x) - tan (x)) sec (x)

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

:Use the basic trigonometric identity

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

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

:Use the basic trigonometric identity

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

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

بسّط:

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

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

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

وحّد الكسور:

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

(a) = a

:احذف الأقواس

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

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

:اضرب كسور

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

(1 - sin (x)) \cdot 1 = 1 - sin (x)

(1 - sin (x)) \cdot 1

(1 - sin (x)) \cdot 1 = (1 - sin (x)) :

اضرب

= (1 - sin (x))

(a) = a

:احذف الأقواس

= 1 - sin (x)

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

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

cos (x) cos (x)

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

:فعّل قانون القوى

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

= cos^{1 + 1} (x)

1 + 1 = 2 :

اجمع الأعداد

= cos^{2} (x)

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

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

بسّط:

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

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

1 - sin^{2} (x)

حلل إلى عوامل:

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

1 - sin^{2} (x)

- 1

قم باخراج العامل المشترك

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

sin^{2} (x) - 1

حلل إلى عوامل:

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

sin^{2} (x) - 1

1^{2}

كـ

1

اكتب مجددًا

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

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

فعّل قانون فرق المربّعات

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

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

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

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

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

اختزل:

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

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

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

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

بسّط

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

1 - sin (x) :

إلغ العوامل المشتركة

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

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

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

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

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

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

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