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العربية

prove sec(θ)-sec(θ)sin^2(θ)=cos(θ)

الحلّ

prove

sec (θ) - sec (θ) sin^{2} (θ) = cos (θ)

صحيح

خطوات الحلّ

sec (θ) - sec (θ) sin^{2} (θ) = cos (θ)

قم بالعمل على الطرف الأيسر

sec (θ) - sec (θ) sin^{2} (θ)

sin, cos :

عبّر بواسطة

sec (θ) - sec (θ) sin^{2} (θ)

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

:Use the basic trigonometric identity

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

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

بسّط:

\frac{1 - sin ^{2} ( θ )}{cos ( θ )}

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

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

\frac{1}{cos ( θ )} sin^{2} (θ)

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

:اضرب كسور

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

1 \cdot sin^{2} (θ) = sin^{2} (θ) :

اضرب

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

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

فعّل القانون

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{cos ( θ )}

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

:فعّل نطريّة فيتاغوروس

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

= \frac{cos ^{2} ( θ )}{cos ( θ )}

cos (θ) :

إلغ العوامل المشتركة

= cos (θ)

= cos (θ)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

p ro v e 2 cos (2 x) = 4 cos^{2} (x) - 2

p ro v e tan (\frac{π}{4} - θ) = \frac{cos ( 2 θ )}{1 + sin ( 2 θ )}

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

p ro v e (csc (α) + cot (α)) (sec (α) - 1) = tan (α)

p ro v e 1 - 2 sin^{2} (θ) = 2 cos^{2} (θ) - 1