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شائع علم المثلثات >

prove tan(θ)+cot(θ)=sec(θ)csc(θ)

الحلّ

prove

tan (θ) + cot (θ) = sec (θ) csc (θ)

الحلّ

صحيح

خطوات الحلّ

tan (θ) + cot (θ) = sec (θ) csc (θ)

قم بالعمل على الطرف الأيسر

tan (θ) + cot (θ)

sin, cos :

عبّر بواسطة

cot (θ) + tan (θ)

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

:Use the basic trigonometric identity

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

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

:Use the basic trigonometric identity

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

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

بسّط:

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

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

sin (θ), cos (θ)

المضاعف المشترك الأصغر لـ:

sin (θ) cos (θ)

sin (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (θ)

cos (θ)

= sin (θ) cos (θ)

اكتب مجددًا الكسور بحيث يكون المقام مشترك

sin (θ) cos (θ)

اضرب كل بسط ومقام بتعبير الذي يؤدّي إلى مقام مشترك

For

\frac{cos ( θ )}{sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

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

For

\frac{sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ)

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

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

= \frac{1}{cos ( θ ) sin ( θ )}

= \frac{1}{cos ( θ ) sin ( θ )}

Rewrite using trig identities

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

:Use the basic trigonometric identity

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

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

:Use the basic trigonometric identity

\frac{1}{\frac{1}{s e c ( θ )} \cdot \frac{1}{c s c ( θ )}}

بسّط

\frac{1}{\frac{1}{s e c ( θ )} \cdot \frac{1}{c s c ( θ )}}

\frac{1}{sec ( θ )} \cdot \frac{1}{csc ( θ )}

اضرب بـ:

\frac{1}{sec ( θ ) csc ( θ )}

\frac{1}{sec ( θ )} \cdot \frac{1}{csc ( θ )}

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

:اضرب كسور

= \frac{1 \cdot 1}{sec ( θ ) csc ( θ )}

1 \cdot 1 = 1 :

اضرب الأعداد

= \frac{1}{sec ( θ ) csc ( θ )}

= \frac{1}{\frac{1}{s e c ( θ ) c s c ( θ )}}

\frac{1}{\frac{b}{c}} = \frac{c}{b}

: استخدم ميزات الكسور التالية

= \frac{sec ( θ ) csc ( θ )}{1}

\frac{a}{1} = a

فعّل القانون

= sec (θ) csc (θ)

sec (θ) csc (θ)

sec (θ) csc (θ)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

أمثلة شائعة

sin(2x)=-(sqrt(3))/2

sin (2 x) = - \frac{3}{2}

sin(2x)= 1/2

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

arctan(4)

arctan (4)

cos(pi/3)

cos (\frac{π}{3})

sinh(1)

sinh (1)