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العربية

prove (sin(a)+tan(a))/(1+cos(a))=tan(a)

الحلّ

prove

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

صحيح

خطوات الحلّ

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

قم بالعمل على الطرف الأيسر

\frac{sin ( a ) + tan ( a )}{1 + cos ( a )}

sin, cos :

عبّر بواسطة

\frac{sin ( a ) + tan ( a )}{1 + cos ( a )}

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

:Use the basic trigonometric identity

= \frac{sin ( a ) + \frac{s i n ( a )}{c o s ( a )}}{1 + cos ( a )}

\frac{sin ( a ) + \frac{s i n ( a )}{c o s ( a )}}{1 + cos ( a )}

بسّط:

\frac{sin ( a )}{cos ( a )}

\frac{sin ( a ) + \frac{s i n ( a )}{c o s ( a )}}{1 + cos ( a )}

sin (a) + \frac{sin ( a )}{cos ( a )}

وحّد:

\frac{sin ( a ) cos ( a ) + sin ( a )}{cos ( a )}

sin (a) + \frac{sin ( a )}{cos ( a )}

sin (a) = \frac{sin ( a ) cos ( a )}{cos ( a )}

:حوّل الأعداد لكسور

= \frac{sin ( a ) cos ( a )}{cos ( a )} + \frac{sin ( a )}{cos ( a )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{sin ( a ) cos ( a ) + sin ( a )}{cos ( a )}

= \frac{\frac{s i n ( a ) c o s ( a ) + s i n ( a )}{c o s ( a )}}{1 + cos ( a )}

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

: استخدم ميزات الكسور التالية

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

sin (a)

قم باخراج العامل المشترك

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

cos (a) + 1 :

إلغ العوامل المشتركة

= \frac{sin ( a )}{cos ( a )}

= \frac{sin ( a )}{cos ( a )}

= \frac{sin ( a )}{cos ( a )}

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

:Use the basic trigonometric identity

= tan (a)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e \frac{4 sec ( x ) - 4}{1 - cos ( θ )} = 4 sec (x)

p ro v e 1 + cot^{2} (x) = \frac{tan ^{2} ( x ) + 1}{tan ^{2} ( x )}

p ro v e sin (18 0^{\circ} + θ) = - sin (θ)

p ro v e \frac{1 - tan ^{2} ( x )}{1 + tan ^{2} ( x )} = sec (x)

p ro v e 2 sec (x) = 2 + 2 tan (x)