prove 1+(tan^2(A))/(1+sec(A))=sec(A)

الحلّ

prove

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )} = sec (A)

صحيح

خطوات الحلّ

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )} = sec (A)

قم بالعمل على الطرف الأيسر

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )}

Rewrite using trig identities

1 + \frac{tan ^{2} ( A )}{1 + sec ( A )}

tan^{2} (x) + 1 = sec^{2} (x)

:فعّل نطريّة فيتاغوروس

tan^{2} (x) = sec^{2} (x) - 1

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

1 + \frac{sec ^{2} ( A ) - 1}{1 + sec ( A )}

بسّط:

sec (A)

1 + \frac{sec ^{2} ( A ) - 1}{1 + sec ( A )}

\frac{sec ^{2} ( A ) - 1}{1 + sec ( A )}

اختزل:

sec (A) - 1

\frac{sec ^{2} ( A ) - 1}{1 + sec ( A )}

sec^{2} (A) - 1

حلل إلى عوامل:

(sec (A) + 1) (sec (A) - 1)

sec^{2} (A) - 1

1^{2}

كـ

1

اكتب مجددًا

= sec^{2} (A) - 1^{2}

x^{2} - y^{2} = (x + y) (x - y)

فعّل قانون فرق المربّعات

sec^{2} (A) - 1^{2} = (sec (A) + 1) (sec (A) - 1)

= (sec (A) + 1) (sec (A) - 1)

= \frac{( sec ( A ) + 1 ) ( sec ( A ) - 1 )}{1 + sec ( A )}

sec (A) + 1 :

إلغ العوامل المشتركة

= sec (A) - 1

= 1 + sec (A) - 1

جمّع التعابير المتشابهة

= sec (A) + 1 - 1

1 - 1 = 0

= sec (A)

= sec (A)

= sec (A)

أظهرنا بأنّ الطرفين من الممكن أن تكون لهما نفس الصورة

\Rightarrow صحيح

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

p ro v e tan (13 5^{\circ} + a) = \frac{tan ( a ) - 1}{tan ( a ) + 1}

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

p ro v e 4 cos^{2} (x) + cos (2 x) = 5 - 6 sin^{2} (x)