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prove ((sec(θ))/(cos(θ)))-((tan(θ))/(cot(θ)))=1

الحلّ

prove

(\frac{sec ( θ )}{cos ( θ )}) - (\frac{tan ( θ )}{cot ( θ )}) = 1

صحيح

خطوات الحلّ

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )} = 1

قم بالعمل على الطرف الأيسر

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

sin, cos :

عبّر بواسطة

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

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

:Use the basic trigonometric identity

= \frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

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

:Use the basic trigonometric identity

= \frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{cot ( θ )}

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

:Use the basic trigonometric identity

= \frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

\frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

بسّط:

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

\frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

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

\frac{\frac{1}{c o s ( θ )}}{cos ( θ )}

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

: استخدم ميزات الكسور التالية

= \frac{1}{cos ( θ ) cos ( θ )}

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

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

:فعّل قانون القوى

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

1 + 1 = 2 :

اجمع الأعداد

= cos^{2} (θ)

= \frac{1}{cos ^{2} ( θ )}

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

\frac{\frac{s i n ( θ )}{c o s ( θ )}}{\frac{c o s ( θ )}{s i n ( θ )}}

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

:اقسم الكسور

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

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

:فعّل قانون القوى

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

1 + 1 = 2 :

اجمع الأعداد

= sin^{2} (θ)

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

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

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

:فعّل قانون القوى

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

1 + 1 = 2 :

اجمع الأعداد

= cos^{2} (θ)

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

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

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

فعّل القانون

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

\frac{a}{a} = 1

فعّل القانون

= 1

= 1

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e sin (π) = sin (3 π)

p ro v e \frac{2 + csc ( x )}{sec ( x )} - 2 cos (x) = cot (x)

p ro v e \frac{cos ( 6 0 ^{\circ} )}{sin ( 6 0 ^{\circ} )} = cot (6 0^{\circ})

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

p ro v e sec (3 0^{\circ}) - sin (3 0^{\circ}) tan (3 0^{\circ}) = cos (3 0^{\circ})