3 tożsamości trygonometryczne

[Danlew]

Witam

proszę o pomoc z tym zadaniem:

Sprawdz czy rownosc jest tozsamoscia trygonometryczna:

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

b) \(\displaystyle{ \frac{\cos \alpha }{1 + \sin \alpha } + \frac{1 + \sin \alpha }{\cos \alpha } = \frac{2}{\cos \alpha }}\)

c) \(\displaystyle{ \cos^{4} \alpha + \sin^{2} \alpha \cdot \cos^{2} \alpha + \sin^{4} \alpha = 1}\)

[Lbubsazob]

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

[Danlew]

Dzieki a nastepne?

[Lbubsazob]

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

[Danlew]

A c?