Question

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Zad 6,33)
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Answer 1

Verified answer

a)

[tex]1+tg^2\alpha=\dfrac{1}{cos^2\alpha}\\\\1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2=\dfrac1{cos^2\alpha}\\\\1+\dfrac{sin^2\alpha}{cos^2\alpha}=\dfrac1{cos^2\alpha}\\\\\dfrac{cos^2\alpha}{cos^2\alpha}+\dfrac{sin^2\alpha}{cos^2\alpha}=\dfrac1{cos^2\alpha}\\\\\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}=\dfrac1{cos^2\alpha}\\\\\dfrac{1}{cos^2\alpha}=\dfrac1{cos^2\alpha}\\\\L=P[/tex]

Równość jest spełniona.

b)

[tex]1+ctg^2\alpha=\dfrac1{sin^2\alpha}\\\\\dfrac{sin^2\alpha}{sin^2\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}=\dfrac1{sin^2\alpha}\\\\\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}=\dfrac1{sin^2\alpha}\\\\L=P[/tex]

Równość jest spełniona.

c)

[tex]sin^4\alpha+cos^2\alpha=sin^2\alpha+cos^4\alpha\\\\sin^4\alpha-cos^4\alpha=sin^2\alpha-cos^2\alpha\\\\(sin^2\alpha-cos^2\alpha)(sin^2\alpha+cos^2\alpha)=sin^2\lpha-cos^2\alpha\\\\(sin^2\alpha-cos^2\alpha)\cdot 1=sin^2\alpha-cos^2\alpha\\\\sin^2\alpha-cos^2\alpha=sin^2\alpha-cos^2\alpha\\\\L=P[/tex]

Równość jest spełniona.

d)

[tex]tg^2\alpha-sin^2\alpha=tg^2\alpha\cdot sin^2\alpha\\\\\dfrac{sin^2\alpha}{cos^2\alpha}-sin^2\alpha=\dfrac{sin^2\alpha}{cos^2\alpha}\cdot sin^2\alpha\\\\\dfrac{sin^2\alpha}{cos^2\alpha}-\dfrac{sin^2\alpha cos^2\alpha}{cos^2\alpha}=\dfrac{sin^4\alpha}{cos^2\alpha}\\\\\dfrac{sin^2\alpha-sin^2\alpha cos^2\alpha}{cos^2\alpha}=\dfrac{sin^4\alpha}{cos^2\alpha}\\\\[/tex]

[tex]\dfrac{sin^2\alpha(1-cos^2\alpha)}{cos^2\alpha}=\dfrac{sin^4\alpha}{cos^2\alpha}\\\\\dfrac{sin^2\alpha\cdot sin^2\alpha}{cos^2\alpha}=\dfrac{sin^4\alpha}{cos^2\alpha}\\\\L=P[/tex]

Równość jest spełniona.

e)

[tex]\dfrac{tg\alpha}{tg\alpha+ctg\alpha}=sin^2\alpha\\\\\dfrac{\frac{sin\alpha}{cos\alpha}}{\frac{sin\alpha}{cos\alpha}+\frac{cos\alpha}{sin\alpha}}=sin^2\alpha\\\\\dfrac{\frac{sin\alpha}{cos\alpha}}{\frac{sin^2\alpha+cos^2\alpha}{sin\alpha cos\alpha}}=sin^2\alpha\\\\\dfrac{sin\alpha}{cos\alpha}:\dfrac1{sin\alpha cos\alpha}=sin^2\alpha\\\\\dfrac{sin\alpha}{cos\alpha}\cdot sin\alpha cos\alpha=sin^2\alpha\\\\\dfrac{sin^2\alpha cos\alpha}{cos\alpha}=sin^2\alpha\\\\sin^2\alpha=sin^2\alpha\\\\L=P[/tex]

Równość jest spełniona.

f)

[tex]\dfrac{sin\alpha+tg\alpha}{1+cos\alpha}=tg\alpha\\\\\dfrac{sin\alpha+\frac{sin\alpha}{cos\alpha}}{1+cos\alpha}=tg\alpha\\\\\dfrac{\frac{sin\alpha cos\alpha+sin\alpha}{cos\alpha}}{1+cos\alpha}=tg\alpha\\\\\dfrac{\frac{sin\alpha(cos\alpha+1)}{cos\alpha}}{1+cos\alpha}=tg\alpha\\\\\dfrac{sin\alpha(cos\alpha+1)}{cos\alpha}\cdot\dfrac1{1+cos\alpha}=tg\alpha\\\\\dfrac{sin\alpha(1+cos\alpha)}{cos\alpha(1+cos\alpha)}=tg\alpha\\\\\dfrac{sin\alpha}{cos\alpha}=tg\alpha\\\\L=P[/tex]

Równość jest spełniona.

g)

[tex]\dfrac{tg\alpha}{1-tg^2\alpha}=\dfrac{ctg\alpha}{ctg^2\alpha-1}\\\\\dfrac{tg\alpha}{1-tg^2\alpha}=\dfrac{\frac1{tg\alpha}}{\left(\frac1{tg\alpha}\right)^2-1}\\\\\dfrac{tg\alpha}{1-tg^2\alpha}=\dfrac{\frac1{tg\alpha}}{\frac{1}{tg^2\alpha}-1}\\\\\dfrac{tg\alpha}{1-tg^2\alpha}=\dfrac1{tg\alpha}:\left(\dfrac{1}{tg^2\alpha}-1\right)\\\\\dfrac{tg\alpha}{1-tg^2\alpha}=\dfrac1{tg\alpha}\cdot\dfrac1{\frac1{tg^2\alpha}-1}\\\\\dfrac{tg\alpha}{1-tg^2\alpha}=\dfrac{1}{\frac1{tg\alpha}-tg\alpha}\\\\[/tex]

[tex]\dfrac{tg\alpha}{1-tg^2\alpha}=\dfrac{tg\alpha}{tg\alpha(\frac{1}{tg\alpha}-tg\alpha)}\\\\\dfrac{tg\alpha}{1-tg^2\alpha}=\dfrac{tg\alpha}{1-tg^2\alpha}\\\\L=P[/tex]

Równość jest spełniona.

h)

[tex]\dfrac1{1+tg^2\alpha}+\dfrac{1}{1+ctg^2\alpha}=1\\\\\dfrac1{1+\frac{sin^2\alpha}{cos^2\alpha}}+\dfrac1{1+\frac{cos^2\alpha}{sin^2\alpha}}=1\\\\\dfrac{1+\frac{cos^2\alpha}{sin^2\alpha}+1+\frac{sin^2\alpha}{cos^2\alpha}}{(1+\frac{sin^2\alpha}{cos^2\alpha})(1+\frac{cos^2\alpha}{sin^2\alpha})}=1\\\\[/tex]

[tex]2+\dfrac{cos^4\alpha+sin^4\alpha}{(sin\alpha cos\alpha)^2}=1+\dfrac{cos^2\alpha}{sin^2\alpha}+\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{sin^2\alpha}{cos^2\alpha}\cdot\dfrac{cos^2\alpha}{sin^2\alpha}\\\\2+\dfrac{cos^4\alpha+sin^4\alpha}{(sin\alpha cos\alpha)^2}=1+\dfrac{cos^4\alpha+sin^4\alpha}{(sin\alpha cos\alpha)^2}+1\\\\L=P[/tex]

Równanie jest spełnione.

i)

[tex]sin\alpha=\sqrt{\dfrac{tg\alpha}{tg\alpha+ctg\alpha}}\\\\sin\alpha=\sqrt{\dfrac{tg\alpha}{\frac{sin\alpha}{cos\alpha}+\frac{cos\alpha}{sin\alpha}}}\\\\sin^2\alpha=\dfrac{tg\alpha}{\frac{sin^2\alpha+cos^2\alpha}{sin\alpha cos\alpha}}\\\\sin^2\alpha=\dfrac{\frac{sin\alpha}{cos\alpha}}{\frac{1}{sin\alpha cos\alpha}}\\\\sin^2\alpha=\dfrac{sin\alpha}{cos\alpha}\cdot sin\alpha cos\alpha\\\\sin^2\alpha=\dfrac{sin^2\alpha cos\alpha}{cos\alpha}\\\\sin^2\alpha = sin^2\alpha\\\\L=P[/tex]

Równość jest spełniona.

j)

[tex]\dfrac{tg^2\alpha-sin^2\alpha}{ctg^2\alpha-cos^2\alpha}=tg^6\alpha\\\\\dfrac{\frac{sin^2\alpha}{cos^2\alpha}-\frac{sin^2\alpha cos^2\alpha}{cos^2\alpha}}{\frac{cos^2\alpha}{sin^2\alpha}-\frac{cos^2\alpha sin^2\alpha}{sin^2\alpha}}=tg^6\alpha\\\\\dfrac{\frac{sin^2\alpha-sin^2\alpha cos^2\alpha}{cos^2\alpha}}{\frac{cos^2\alpha-cos^2\alpha sin^2\alpha}{sin^2\alpha}}=tg^6\alpha\\\\\dfrac{sin^2\alpha(1-cos^2\alpha)}{cos^2\alpha}:\dfrac{cos^2\alpha(1-sin^2\alpha)}{sin^2\alpha}=tg^6\alpha\\\\[/tex]

[tex]\dfrac{sin^2\alpha(1-cos^2\alpha)}{cos^2\alpha}\cdot\dfrac{sin^2\alpha}{cos^2\alpha(1-sin^2\alpha)}=tg^6\alpha\\\\\dfrac{sin^4\alpha(1-cos^2\alpha)}{cos^4\alpha(1-sin^2\alpha)}=tg^6\alpha\\\\\dfrac{sin^4\alpha\cdot sin^2\alpha}{cos^4\alpha \cdot cos^2\alpha}=tg^6\alpha\\\\\dfrac{sin^6\alpha}{cos^6\alpha}=tg^6\alpha\\\\L=P[/tex]

Równość jest spełniona.