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Answer 1

@Michele_Laino

Answer 2

Is this like the one we did?

Answer 3

we have to start, from the left side, for example. So we can write:
\[\Large \frac{{\csc \theta - 1}}{{\cot \theta }} = \frac{{\frac{1}{{\sin \theta }} - 1}}{{\frac{{\cos \theta }}{{\sin \theta }}}}\]

Answer 4

ther, I have used these identities:
\[\Large \csc \theta = \frac{1}{{\sin \theta }},\quad \cot \theta = \frac{{\cos \theta }}{{\sin \theta }}\]

Answer 5

oops..there*

Answer 6

why would we use cos/sin instead of 1/cot

Answer 7

I mean 1/tan

Answer 8

So it is easier.

Answer 9

in order to get the right side

Answer 10

Do I cross out the sin's for the next step

Answer 11

for next step we can sum these fractions:
\[\Large \frac{1}{{\sin \theta }} - 1 = \frac{1}{{\sin \theta }} - \frac{1}{1} = \frac{{1 - \sin \theta }}{{\sin \theta }}\]

Answer 12

Got it, I dont know the next step

Answer 13

The next step is to evaluate this

Answer 14

But there's no cot on the righthand side

Answer 15

the next step is to compute the subsequent division:
\[\Large \frac{{\frac{{1 - \sin \theta }}{{\sin \theta }}}}{{\frac{{\cos \theta }}{{\sin \theta }}}} = \frac{{1 - \sin \theta }}{{\sin \theta }} \times \frac{{\sin \theta }}{{\cos \theta }} = ...?\]

Answer 16

Oh I see I forgot cos/sin was in the problem

Answer 17

so we get:
\[\Large \frac{{1 - \sin \theta }}{{\sin \theta }} \times \frac{{\sin \theta }}{{\cos \theta }} = \frac{{1 - \sin \theta }}{{\cos \theta }}\]

Answer 18

Got that now

Answer 19

now, I multiply both numerator and denominator by cos(theta):
\[\frac{{1 - \sin \theta }}{{\cos \theta }} \cdot \frac{{\cos \theta }}{{\cos \theta }} = \frac{{\cos \theta \left( {1 - \sin \theta } \right)}}{{{{\left( {\cos \theta } \right)}^2}}}\]

Answer 20

would that change back to (1-sin)/cos?

Answer 21

And I have in about an hour for school

Answer 22

no, since now I substitute cos^2 with 1-sin^2

Answer 23

*And I have about an hour before I have to go to school

Answer 24

like below:
\[\begin{gathered}
\frac{{1 - \sin \theta }}{{\cos \theta }} \cdot \frac{{\cos \theta }}{{\cos \theta }} = \frac{{\cos \theta \left( {1 - \sin \theta } \right)}}{{{{\left( {\cos \theta } \right)}^2}}} = \frac{{\cos \theta \left( {1 - \sin \theta } \right)}}{{1 - {{\left( {\sin \theta } \right)}^2}}} = \hfill \\
\hfill \\
= \frac{{\cos \theta \left( {1 - \sin \theta } \right)}}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}} \hfill \\
\end{gathered} \]

Answer 25

since we have:
\[1 - {\left( {\sin \theta } \right)^2} = \left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)\]

Answer 26

please, keep in mind that the subsequent algebraic identity holds:
a^2-b^2=(a-b)(a+b)

Answer 27

where a^2= 1, and b^2=sin^2, so a=1 and b= sin(theta)

Answer 28

Do I cross out the 1-sin's next?

Answer 29

yes!

Answer 30

so we get:
\[\frac{{\cos \theta \left( {1 - \sin \theta } \right)}}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}} = \frac{{\cos \theta }}{{1 + \sin \theta }}\]

Answer 31

then what happens?

Answer 32

now we have to divide both numerator and denominator by sin(theta), like below:

Answer 33

\[\Large \frac{{\cos \theta }}{{1 + \sin \theta }} = \frac{{\frac{{\cos \theta }}{{\sin \theta }}}}{{\frac{{1 + \sin \theta }}{{\sin \theta }}}}\]

Answer 34

Why do we have to divide it by sin?

Answer 35

because, in so doing we get the right side of your identity

Answer 36

please note that, at the numerator we have:
\[\frac{{\cos \theta }}{{\sin \theta }} = \cot \theta \]

Answer 37

whereas at the denominator, we have:
\[\frac{{1 + \sin \theta }}{{\sin \theta }} = \frac{1}{{\sin \theta }} + \frac{{\sin \theta }}{{\sin \theta }} = \frac{1}{{\sin \theta }} + 1\]

Answer 38

namely:
\[\frac{1}{{\sin \theta }} + 1 = \csc \theta + 1\]

Answer 39

Where does the plus one come from?

Answer 40

since we have:
\[\frac{{1 + \sin \theta }}{{\sin \theta }} = \frac{1}{{\sin \theta }} + \frac{{\sin \theta }}{{\sin \theta }}\]

Answer 41

and:

Answer 42

I still don't see where it comes from shouldn't it just be 1/sin?

Answer 43

for example if I have to divide this:
\[\frac{{5 + 1}}{5}\]

Answer 44

then I can write:
\[\frac{{5 + 1}}{5} = \frac{5}{5} + \frac{1}{5} = 1 + \frac{1}{5}\]

Answer 45

Ohhh got it

Answer 46

so, we get:
\[\frac{{\frac{{\cos \theta }}{{\sin \theta }}}}{{\frac{{1 + \sin \theta }}{{\sin \theta }}}} = \frac{{\cot \theta }}{{\csc \theta + 1}}\]

Answer 47

\[\Large \frac{{\frac{{\cos \theta }}{{\sin \theta }}}}{{\frac{{1 + \sin \theta }}{{\sin \theta }}}} = \frac{{\cot \theta }}{{\csc \theta + 1}}\]

Answer 48

namely, the right side of your identity

Answer 49

Thank you, that's all I can do for right now because I have to go to school, do you know when you'll 've on later

Answer 50

I will be here from 17:00 to 22:00 (Italy time zone)

Answer 51

Umm I'll be back in nine hours

Answer 52

ok!

Answer 53

then I will be here from 17:00 to 24:00 (Italy time zone)