Sprawdź czy podane równości są tożsamościami tryg.

emilianov · Post autor: **emilianov** » 9 maja 2012, o 18:28

jw + nalezy podac zalozenia:

1) \(\displaystyle{ \frac{\sin x + \cos x}{\cos x} = 1 + \tg x}\)

2) \(\displaystyle{ \frac{\cos x + \ctg x}{\cos x} = 1 + \frac{1}{\sin x}}\)

3) \(\displaystyle{ \frac{\tg x(1 + \ctg ^{2}x)}{1 + \tg ^{2}x} = \ctg x}\)

Lbubsazob · Post autor: **Lbubsazob** » 9 maja 2012, o 18:36

1) \(\displaystyle{ \frac{\sin x+\cos x}{\cos x}= \frac{\sin x}{\cos x}+ \frac{\cos x}{\cos x}= \frac{\sin x}{\cos x}+1=\tg x+1}\)

2) \(\displaystyle{ \frac{\cos x+\ctg x}{\cos x}= \frac{\cos x}{\cos x}+ \frac{\ctg x}{\cos x}=1+ \frac{ \frac{\cos x}{\sin x} }{\cos x}=1+ \frac{1}{\sin x}}\)

3) \(\displaystyle{ \frac{\tg x\left( 1+\ctg^2x\right) }{1+\tg^2x}= \frac{\tg x+\tg x\ctg^2x}{1+\tg^2x}= \frac{\tg x+\ctg x}{1+\tg^2x}= \frac{ \tg x+ \frac{1}{\tg x} }{ 1+\tg^2x }= \frac{\frac{\tg^2x+1}{\tg x} }{1+\tg^2x}= \frac{1}{\tg x}=\ctg x}\)

DonArkanio · Post autor: **DonArkanio** » 9 maja 2012, o 18:41

Oj

1)
\(\displaystyle{ \frac{\sin x}{\cos x}+ \frac{\cos x}{\cos x}= 1+ \tg x}\)

z tego

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

\(\displaystyle{ \frac{\sin x}{\cos x} = \tg x}\)

L=P

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