Proszę o pomoc w przykładach g,h,i,j. Nie potrzebuję wszystkich, uciesze się z każdej pomocy.... Question from @Zulimen1111

January 2019 1 11 Report

Proszę o pomoc w przykładach g,h,i,j. Nie potrzebuję wszystkich, uciesze się z każdej pomocy.

ghe F)
$\frac{1+cos\alpha}{sin\alpha}=\frac{sin\alpha}{1-cos\alpha}$

$L=\frac{1+cos\alpha}{sin\alpha}=$

$\frac{(1+cos\alpha)(1-cos\alpha)}{sin\alpha(1-cos\alpha)}=$

$\frac{1-cos^2\alpha}{sin\alpha(1-cos\alpha)}=$

$\frac{sin^2\alpha}{sin\alpha(1-cos\alpha)}=$

$\frac{sin\alpha}{1-cos\alpha}=P$
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g)
$\left(\frac{1}{cos\alpha}-\frac{1}{sin\alpha}\right)\cdot(1+tg\alpha+ctg\alpha)=\frac{sin\alpha}{cos^2\alpha}-\frac{cos\alpha}{sin^2\alpha}$

$L=\left(\frac{1}{cos\alpha}-\frac{1}{sin\alpha}\right)\cdot(1+tg\alpha+ctg\alpha)=$

$\left(\frac{1}{cos\alpha}-\frac{1}{sin\alpha}\right)\cdot\left(1+ \frac{sin\alpha}{cos\alpha} + \frac{cos\alpha}{sin\alpha}\right) =$

$\frac{1}{cos\alpha}+ \frac{sin\alpha}{cos^2\alpha}+\frac{1}{sin\alpha}- \frac{1}{sin\alpha}- \frac{1}{cos\alpha} - \frac{cos\alpha}{sin^2\alpha}=$

$\frac{sin\alpha}{cos^2\alpha}-\frac{cos\alpha}{sin^2\alpha}=P$
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h)
$\left(\frac{1}{cos\alpha}+\frac{1}{sin\alpha}\right)\cdot(tg\alpha+ctg\alpha-1)=\frac{sin\alpha}{cos^2\alpha}+\frac{cos\alpha}{sin^2\alpha}$

$L=\left(\frac{1}{cos\alpha}+\frac{1}{sin\alpha}\right)\cdot(tg\alpha+ctg\alpha-1)=$

$\left(\frac{1}{cos\alpha}+\frac{1}{sin\alpha}\right)\cdot\left( \frac{sin\alpha}{cos\alpha} + \frac{cos\alpha}{sin\alpha} -1\right) =$

$\frac{sin\alpha}{cos^2\alpha}+\frac{1}{sin\alpha}- \frac{1}{cos\alpha}+\frac{1}{cos\alpha} + \frac{cos\alpha}{sin^2\alpha} - \frac{1}{sin\alpha} =$

$\frac{sin\alpha}{cos^2\alpha}+\frac{cos\alpha}{sin^2\alpha}=P$
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i)
$(1+sin\alpha)\cdot\left(\frac{1}{cos\alpha}-\frac{1}{ctg\alpha}\right)-cos\alpha=0$

$L=(1+sin\alpha)\cdot\left(\frac{1}{cos\alpha}-\frac{1}{ctg\alpha}\right)-cos\alpha=$

$(1+sin\alpha)\cdot\left(\frac{1}{cos\alpha}-\frac{1}{ \frac{cos\alpha}{sin\alpha} }\right)-cos\alpha=$

$(1+sin\alpha)\cdot\left(\frac{1}{cos\alpha}- \frac{sin\alpha}{cos\alpha }\right)-cos\alpha=$

$\frac{1}{cos\alpha}- \frac{sin\alpha}{cos\alpha}+ \frac{sin\alpha}{cos\alpha }- \frac{sin^2\alpha}{cos\alpha} -cos\alpha=$

$\frac{1}{cos\alpha}- \frac{sin^2\alpha}{cos\alpha} -cos\alpha=$

$\frac{1-sin^2\alpha}{cos\alpha} -cos\alpha=$

$\frac{cos^2\alpha}{cos\alpha} -cos\alpha=$

$cos\alpha -cos\alpha=0=P$
==================
j)
$(1-cos\alpha)\cdot\left(\frac{1}{sin\alpha}+\frac{1}{tg\alpha}\right)-sin\alpha=0$

$L=(1-cos\alpha)\cdot\left(\frac{1}{sin\alpha}+\frac{1}{tg\alpha}\right)-sin\alpha=$

$(1-cos\alpha)\cdot\left(\frac{1}{sin\alpha}+\frac{1}{ \frac{sin\alpha}{cos\alpha} }\right)-sin\alpha=$

$(1-cos\alpha)\cdot\left(\frac{1}{sin\alpha}+\frac{cos\alpha}{sin\alpha}\right)-sin\alpha=$

$\frac{1}{sin\alpha}+\frac{cos\alpha}{sin\alpha}- \frac{cos\alpha}{sin\alpha}- \frac{cos^2\alpha}{sin\alpha} -sin\alpha=$

$\frac{1}{sin\alpha}- \frac{cos^2\alpha}{sin\alpha} -sin\alpha=$

$\frac{1- cos^2\alpha}{sin\alpha} -sin\alpha=$

$\frac{sin^2\alpha}{sin\alpha} -sin\alpha=$

$sin\alpha -sin\alpha=0=P$

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