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العربية

prove (1-sin(θ))/(sec(θ))=(cos^3(θ))/(1+sin(θ))

الحلّ

prove

\frac{1 - sin ( θ )}{sec ( θ )} = \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

صحيح

خطوات الحلّ

\frac{1 - sin ( θ )}{sec ( θ )} = \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

قم بالعمل على الطرف الأيسر

\frac{1 - sin ( θ )}{sec ( θ )}

Rewrite using trig identities

\frac{1 + sin ( θ )}{1 + sin ( θ )}

اضرب بـ

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

(1 + sin (θ)) (1 - sin (θ))

وسٌع:

1 - sin^{2} (θ)

(1 + sin (θ)) (1 - sin (θ))

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

فعّل قانون فرق المربّعات

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

1^{a} = 1

فعّل القانون

1^{2} = 1

= 1 - sin^{2} (θ)

= \frac{1 - sin ^{2} ( θ )}{( 1 + sin ( θ ) ) sec ( θ )}

1 = cos^{2} (θ) + sin^{2} (θ)

:فعّل نطريّة فيتاغوروس

1 - sin^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{( 1 + sin ( θ ) ) sec ( θ )}

= \frac{cos ^{2} ( θ )}{( 1 + sin ( θ ) ) sec ( θ )}

sin, cos :

عبّر بواسطة

\frac{cos ^{2} ( θ )}{( 1 + sin ( θ ) ) sec ( θ )}

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

:Use the basic trigonometric identity

= \frac{cos ^{2} ( θ )}{( 1 + sin ( θ ) ) \frac{1}{c o s ( θ )}}

\frac{cos ^{2} ( θ )}{( 1 + sin ( θ ) ) \frac{1}{c o s ( θ )}}

بسّط:

\frac{cos ^{3} ( θ )}{1 + sin ( θ )}

\frac{cos ^{2} ( θ )}{( 1 + sin ( θ ) ) \frac{1}{c o s ( θ )}}

(1 + sin (θ)) \frac{1}{cos ( θ )}

اضرب بـ:

\frac{1 + sin ( θ )}{cos ( θ )}

(1 + sin (θ)) \frac{1}{cos ( θ )}

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

:اضرب كسور

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

1 \cdot (1 + sin (θ)) = 1 + sin (θ)

1 \cdot (1 + sin (θ))

1 \cdot (1 + sin (θ)) = (1 + sin (θ)) :

اضرب

= (1 + sin (θ))

(a) = a

:احذف الأقواس

= 1 + sin (θ)

= \frac{1 + sin ( θ )}{cos ( θ )}

= \frac{cos ^{2} ( θ )}{\frac{1 + s i n ( θ )}{c o s ( θ )}}

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

: استخدم ميزات الكسور التالية

= \frac{cos ^{2} ( θ ) cos ( θ )}{1 + sin ( θ )}

cos^{2} (θ) cos (θ) = cos^{3} (θ)

cos^{2} (θ) cos (θ)

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

:فعّل قانون القوى

cos^{2} (θ) cos (θ) = cos^{2 + 1} (θ)

= cos^{2 + 1} (θ)

2 + 1 = 3 :

اجمع الأعداد

= cos^{3} (θ)

= \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

= \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

= \frac{cos ^{3} ( θ )}{1 + sin ( θ )}

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e 5 sin^{2} (x) + 5 cos^{2} (x) = 5

p ro v e cot (θ) = \frac{( 1 + cos ( 2 θ ) )}{sin ( 2 θ )}

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

p ro v e tan^{2} (2 3^{\circ}) - sec^{2} (2 3^{\circ}) = - 1

p ro v e (1 - cos^{2} (t)) (1 + cos^{2} (t)) = 1