Question

(csc-1)(csc+1)=cot^2
csc^2-1=cot^2
cot^2=cot^2

Answer 1

@Michele_Laino is this correct?

Answer 2

that's right!

Answer 3

Awesome, thank you, how much longer will you be on?

Answer 4

I stay here, for at least 2 hours

Answer 5

\[\cos^2\theta(1+\tan ^2\theta)=1\]
\[\cos^2\theta(\sec^2\theta)=1\]

\[(\cos^2\theta/1)(1/\cos^2\theta)=1\]
1=1

Answer 6

Yeah! That's correct!

Answer 7

that's right!

Answer 8

\[\sec \theta - \tan \theta= \cos \theta/1+\sin \theta\]


\[(1/\cos)-(\sin/\cos)= \cos/1+\sin\]

Then would the next line by 1-sin

Answer 9

yes! it would be:
\[\frac{{1 - \sin }}{{\cos }}\]

Answer 10

the cos wouldn't cancel?

Answer 11

no, since the right side is:
\[\frac{{\cos }}{{1 + \sin }}\]

Answer 12

we can use this identity:
\[\cos = \sqrt {1 - {{\sin }^2}} = \sqrt {1 - \sin } \sqrt {1 + \sin } \]

Answer 13

Is there another identity that can be used? Our teacher doesnt have that one as an identity we can use

Answer 14

for example, the left side, will be like below:
\[\large \frac{{1 - \sin }}{{\cos }} = \frac{{1 - \sin }}{{\sqrt {1 - \sin } \times \sqrt {1 + \sin } }} = \frac{{\sqrt {1 - \sin } \times \sqrt {1 - \sin } }}{{\sqrt {1 - \sin } \times \sqrt {1 + \sin } }} = ...?\]

Answer 15

please, wait, I'm searching for another method

Answer 16

here it is another method:
\[\frac{{1 - \sin }}{{\cos }} = \frac{{1 - \sin }}{{\cos }} \cdot \frac{{\cos }}{{\cos }} = \frac{{\cos \left( {1 - \sin } \right)}}{{{{\cos }^2}}} = \frac{{\cos \left( {1 - \sin } \right)}}{{1 - {{\sin }^2}}} = \frac{{\cos \left( {1 - \sin } \right)}}{{\left( {1 - \sin } \right)\left( {1 + \sin } \right)}} = ...?\]

Answer 17

\[\begin{gathered}
\frac{{1 - \sin }}{{\cos }} = \frac{{1 - \sin }}{{\cos }} \cdot \frac{{\cos }}{{\cos }} = \frac{{\cos \left( {1 - \sin } \right)}}{{{{\cos }^2}}} = \frac{{\cos \left( {1 - \sin } \right)}}{{1 - {{\sin }^2}}} = \hfill \\
= \frac{{\cos \left( {1 - \sin } \right)}}{{\left( {1 - \sin } \right)\left( {1 + \sin } \right)}} = ...? \hfill \\
\end{gathered} \]

Answer 18

namely I have multiplied and divided, the left side by cos(theta)

Answer 19

so we can write:

Answer 20

Why didn't distribute the cos?

Answer 21

because we have to solve an identity, so we have to start from the left side, and using some computation, we have to transform that left side into the right side

Answer 22

or, similarly starting from the right side, we have to transform that right side into the left side

Answer 23

And what happened to the cos^2

Answer 24

I can rewrite cos^2 as below:
cos^2 = 1-sin^2 =(1-sin)(1+sin)
namely I have used the fundamental identity

Answer 25

why wouldnt it be two (1-sin)?

Answer 26

they cancel each other, since one factor (1-sin) is at the numerator, and the second one is at denominator

Answer 27

there I have applied this algebraic identity:

a^2-b^2= (a-b)(a+b)

Answer 28

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

Answer 29

Okay I understand

Answer 30

\[9\sec^2\theta-5\tan^2\theta=5+4\sec^2\theta\]

Answer 31

here, we can start from the left side, and we can write this:
\[\Large \begin{gathered}
9{\left( {\sec \theta } \right)^2} - 5{\left( {\tan \theta } \right)^2} = \hfill \\
\\
=9\frac{1}{{{{\left( {\cos \theta } \right)}^2}}} - 5\frac{{{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} = \frac{{9 - 5{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} = ...? \hfill \\
\end{gathered} \]

Answer 32

I dont know what to do next

Answer 33

hint:
\[\Large \begin{gathered}
\frac{{9 - 5{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} = \frac{{4 + 5 - 5{{\left( {\sin \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} = \hfill \\
\hfill \\
= \frac{{4 + 5\left( {1 - {{\left( {\sin \theta } \right)}^2}} \right)}}{{{{\left( {\cos \theta } \right)}^2}}} = \frac{{4 + 5{{\left( {\cos \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} = ...? \hfill \\
\end{gathered} \]

Answer 34

Where did the -5 go?

Answer 35

I have factored out 5, so I could write:
\[\Large 5 - 5{\left( {\sin \theta } \right)^2} = 5\left( {1 - {{\left( {\sin \theta } \right)}^2}} \right)\]

Answer 36

I understand that now, then what happens next?

Answer 37

we have the subsequent steps:
\[\Large \frac{{4 + 5{{\left( {\cos \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} = \frac{4}{{{{\left( {\cos \theta } \right)}^2}}} + \frac{{5{{\left( {\cos \theta } \right)}^2}}}{{{{\left( {\cos \theta } \right)}^2}}} = \frac{4}{{{{\left( {\cos \theta } \right)}^2}}} + 5\]

Answer 38

Got that part now

Answer 39

Do we change the cos next?

Answer 40

we can use this identity:
\[\Large \frac{1}{{{{\left( {\cos \theta } \right)}^2}}} = {\left( {\sec \theta } \right)^2}\]

Answer 41

so we get:
\[\Large \frac{4}{{{{\left( {\cos \theta } \right)}^2}}} + 5 = 4{\left( {\sec \theta } \right)^2} + 5\]

Answer 42

Got it

Answer 43

Sorry its so sloppy, I didnt feel like typing it out

Answer 44

that's right!

Answer 45

your method is right!

Answer 46

Would I multiple the cot to the 1, and the 1 on top of cos

Answer 47

here is the next step:
\[\Large \frac{{1 + \tan }}{{1 - \tan }} = \frac{{1 + \frac{1}{{\cot }}}}{{1 - \frac{1}{{\cot }}}} = \frac{{\frac{{\cot + 1}}{{\cot }}}}{{\frac{{\cot - 1}}{{\cot }}}} = \frac{{\cot + 1}}{{\cot }} \times \frac{{\cot }}{{\cot - 1}} = ...?\]

Answer 48

\[\Large \begin{gathered}
\frac{{1 + \tan }}{{1 - \tan }} = \frac{{1 + \frac{1}{{\cot }}}}{{1 - \frac{1}{{\cot }}}} = \frac{{\frac{{\cot + 1}}{{\cot }}}}{{\frac{{\cot - 1}}{{\cot }}}} = \hfill \\
\hfill \\
= \frac{{\cot + 1}}{{\cot }} \times \frac{{\cot }}{{\cot - 1}} = ...? \hfill \\
\end{gathered} \]

Answer 49

Did you multiple cot to something to get cot+1

Answer 50

no, since I have computed this:
\[\Large 1 + \tan = 1 + \frac{1}{{\cot }} = \frac{1}{1} + \frac{1}{{\cot }} = \frac{{1 \cdot \cot + 1}}{{\cot }} = \frac{{1 + \cot }}{{\cot }}\]

Answer 51

namely I have summed the 2 fraction 1 and 1/cot each other

Answer 52

fractions*

Answer 53

similarly for:
1-tan

Answer 54

I still dont understand sorry

Answer 55

no worries, I retry, please wait a moment

Answer 56

we have to sum this quantities:
1=1/1 and 1/cot(theta)
Now the least common multiple between 1 and cot(theta) is:
cot(theta)

Answer 57

yes

Answer 58

so we have:
\[\Large 1 + \tan = 1 + \frac{1}{{\cot }} = \frac{1}{1} + \frac{1}{{\cot }} = \frac{{1 \cdot \cot + 1}}{{\cot }}\]

Answer 59

\[\Large \frac{1}{1} + \frac{1}{{\cot }} = \frac{{1 \cdot \cot + 1}}{{\cot }}\]

Answer 60

for example if I have to sum these fractions:
1 and 1/5, then the least common multiple is 5, so Ican write:
\[\Large 1 + \frac{1}{5} = \frac{1}{1} + \frac{1}{5} = \frac{{1 \times 5 + 1}}{5}\]

Answer 61

oops..so I can write...

Answer 62

Ohhh I get it now

Answer 63

do you have to go?

Answer 64

no, I stay here

Answer 65

okay, I'm going to start a new post because this one is getting too long

Answer 66

ok!