Tożsamość trygonometryczna do rozwiązania

[sulik7]

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


Jak to rozwiązać?? Mam sprawdzić tożsamość.

[soku11]

\(\displaystyle{ \alpha=x\\
L=\frac{tgx\cdot (1+ctg^2x)}{1+\frac{1}{ctg^2x}}=
\frac{tgx\cdot (1+ctg^2x)}{\frac{1+ctg^2x}{ctg^2x}}=
tgx\cdot ctg^2x=\frac{ctg^2x}{ctgx}=ctgx=P\\
C.N.D}\)

POZDRO

[Jestemfajny]

\(\displaystyle{ \frac{tg\alpha (1+ ctg ^ 2 )}{1+tg^2\alpha} = ctg\alpha \ \ \|*(1+tg^{2} ) \\
tg\alpha(1+ctg^{2}\alpha)=ctg\alpha(1+tg^{2}\alpha)\\
tg\alpha(1+ctg^{2}\alpha)=\frac{1}{tg\alpha}(1+tg^{2}\alpha) \ \ |*tg\alpha \\
tg^{2}\alpha(1+ctg^{2}\alpha)=1+tg^{2}\alpha\\
tg^{2}\alpha +(tg\alpha ctg\alpha)^{2}=1+tg^{2}\alpha\\
1+tg^{2}=1+tg^{2}}\)

[mihal89]

\(\displaystyle{ L = \frac{\frac{sin\alpha}{cos\alpha}(1 + \frac{cos^{2}\alpha}{sin^{2}\alpha})}{1 + \frac{sin^{2}\alpha}{cos^{2}\alpha}} =
\frac{\frac{sin\alpha}{cos\alpha}(\frac{sin^{2}\alpha + cos^{2}\alpha}{sin^{2}\alpha})}{\frac{cos^{2}\alpha + sin^{2}\alpha}{cos^{2}\alpha}} =
\frac{\frac{sin\alpha}{cos\alpha}\frac{1}{sin^{2}\alpha}}{\frac{1}{cos^{2}\alpha}} =
\frac{\frac{1}{sin\alpha cos\alpha}}{\frac{1}{cos^{2}\alpha}} = \\
= \frac{1}{sin\alpha cos\alpha} cos^{2}\alpha = \frac{cos\alpha}{sin\alpha} = ctg\alpha}\)