Question

Can anyone help me prove (7csc(X)-7)/cot(x)=(7cot(x))/(csc(x)+1)

Answer 1

I can rewrite the left side like below:
\[\Large \frac{{7\csc x - 7}}{{\cot x}} = \frac{{\frac{7}{{\sin x}} - 7}}{{\frac{{\cos x}}{{\sin x}}}}\]

Answer 2

is it ok?

Answer 3

Yes

Answer 4

now, we have:
\[\Large \frac{7}{{\sin x}} - 7 = \frac{{7 - 7\sin x}}{{\sin x}}\]

Answer 5

Now I'm lost. Where did the cos(x) go?

Answer 6

that is the numerator only

Answer 7

Okay that makes more sense

Answer 8

so, we can write:
\[\Large \begin{gathered}
\frac{{\frac{7}{{\sin x}} - 7}}{{\frac{{\cos x}}{{\sin x}}}} = \left( {\frac{7}{{\sin x}} - 7} \right) \times \frac{{\sin x}}{{\cos x}} = \hfill \\
\hfill \\
= \frac{{7 - 7\sin x}}{{\sin x}} \times \frac{{\sin x}}{{\cos x}} = ...? \hfill \\
\end{gathered} \]

Answer 9

\[\frac{ 7-7sinx }{ cosx } ?\]

Answer 10

that's right!

Answer 11

and we can rewrite it as below:
\[\Large \frac{{7\left( {1 - \sin x} \right)}}{{\sin x}}\]

Answer 12

is it ok?

Answer 13

Yes

Answer 14

oops..the right expression is:

\[\frac{{7\left( {1 - \sin x} \right)}}{{\cos x}}\]

Answer 15

Now, I multiply numerator and denominator by cosx, so I get:
\[\Large \frac{{7\left( {1 - \sin x} \right)}}{{\cos x}} \times \frac{{\cos x}}{{\cos x}} = \frac{{7\left( {1 - \sin x} \right)\cos x}}{{{{\left( {\cos x} \right)}^2}}}\]

Answer 16

Then we can use the Pyth. ID right?

Answer 17

that's right!

Answer 18

and we get this:
\[\Large \begin{gathered}
\frac{{7\left( {1 - \sin x} \right)\cos x}}{{{{\left( {\cos x} \right)}^2}}} = \frac{{7\left( {1 - \sin x} \right)\cos x}}{{1 - {{\left( {\sin x} \right)}^2}}} = \hfill \\
\hfill \\
= \frac{{7\left( {1 - \sin x} \right)\cos x}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}} \hfill \\
\end{gathered} \]

Answer 19

Then the 1-sinx cancel out

Answer 20

yes!

Answer 21

you should get this:
\[\Large \frac{{7\left( {1 - \sin x} \right)\cos x}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}} = \frac{{7\cos x}}{{1 + \sin x}}\]

Answer 22

Yes

Answer 23

ok! Now I divide both numerator and denominator by sinx, and I get:
\[\Large \frac{{7\cos x}}{{1 + \sin x}} = \frac{{7\frac{{\cos x}}{{\sin x}}}}{{\frac{{1 + \sin x}}{{\sin x}}}}\]

Answer 24

ok?

Answer 25

So the numerator would be 7cotx

Answer 26

that's right!

Answer 27

whereas the denominator can be rewritten as below:

Answer 28

\[\Large \begin{gathered}
\frac{{1 + \sin x}}{{\sin x}} = \frac{1}{{\sin x}} + \frac{{\sin x}}{{\sin x}} = \hfill \\
\hfill \\
= \frac{1}{{\sin x}} + 1 = \csc x + 1 \hfill \\
\hfill \\
\end{gathered} \]

Answer 29

That's what I got. Thank you so much

Answer 30

Thank you!

Answer 31

by definitions
1/sin x= csc x
cos x/sin x = cot x
so [(7/sinx) -7]/(cosx/sin x)= (7csc x -7)/cot (x)