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العربية

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

الحلّ

prove

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

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيمن

sec (x) + tan (x)

sin, cos :

عبّر بواسطة

sec (x) + tan (x)

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

:Use the basic trigonometric identity

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

بسّط:

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

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

Rewrite using trig identities

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

اضرب بـ

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

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

:فعّل نطريّة فيتاغوروس

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

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

cos (x) :

إلغ العوامل المشتركة

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

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

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

arccsc (- 2)

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

cos (22 5^{\circ})

arctan (\frac{5}{2}) \cdot \frac{180}{π}

cos (3 x) = \frac{1}{2}