Odpowiedź:
[tex]\frac{1 + 2 sin x *cos x}{cos^{2}x } = ( 1 + tg x )^{2}[/tex]
[tex]P = 1 + 2 tg x + tg^{2} x = 1 + 2*\frac{sin x}{cos x} + ( \frac{sin^2x}{cos^2 x} ) =[/tex]
= [tex]\frac{cos^2x}{cos^2 x} + 2*\frac{sin x *cos x}{cos^2 x} + \frac{sin^2x}{cos^2 x} =[/tex] [tex]\frac{cos^2 + sin^2 + 2 sin x*cos x}{cos^2 x} = \frac{1 + 2 sin x* cos x}{cos^2 x} = L[/tex]
Szczegółowe wyjaśnienie:
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Odpowiedź:
[tex]\frac{1 + 2 sin x *cos x}{cos^{2}x } = ( 1 + tg x )^{2}[/tex]
[tex]P = 1 + 2 tg x + tg^{2} x = 1 + 2*\frac{sin x}{cos x} + ( \frac{sin^2x}{cos^2 x} ) =[/tex]
= [tex]\frac{cos^2x}{cos^2 x} + 2*\frac{sin x *cos x}{cos^2 x} + \frac{sin^2x}{cos^2 x} =[/tex] [tex]\frac{cos^2 + sin^2 + 2 sin x*cos x}{cos^2 x} = \frac{1 + 2 sin x* cos x}{cos^2 x} = L[/tex]
Szczegółowe wyjaśnienie: