Question

Trigo help..

Answer 1

\[\frac{ \tan (A) }{ 1 - \cot(A) } + \frac{ \cot (A) }{ 1 - \tan(A) } = \sec(A)cosec(A) + 1\]

Answer 2

@Michele_Laino

Answer 3

@welshfella

Answer 4

this one is tricky - i tried converting everything to sine and cosines but it only became more complicated.

Answer 5

it miggt help to convert all the cot A to tan A

Answer 6

- that doesn't help

Answer 7

@mathslover

Answer 8

@paki @IrishBoy123 @welshfella

Answer 9

please try this substitution:
\[\cot A = \frac{1}{{\tan A}}\]

Answer 10

thats what I tried but didnt get very far
- I'll have another try

Answer 11

It's getting complicated >.<

Answer 12

@Michele_Laino

Answer 13

i boiled it down to t^3 - 1
------
t(t - 1)


where t = tan A

Answer 14

(Tan A/1-Cot A) +( Cot A / 1 - Tan A )= Sec A.Cosec A + 1
= (sin A/cos A) / [1- (cos A/sin A)] + (cos A/sin A) / [1- (sin A/cos A)]

= (sin A/cos A) / [(sin A - cos A)/sin A] + (cos A/sin A) / [(cos A - sin A) /cos A]

= (cos^2 A) / [sin A(cos A - sin A)] - (sin^2 A) / [cos A(cos A - sin A)]

= (cos^3 A - sin^3 A) / [(cos A * sin A) (cos A - sin A)]

= (1/cos A * sin A) (sin^2 A + cos^2 A + sin A cos A)

= (1/cos A * sin A) (1+ sin A cos A)

= [1 / (cos A sin A) + (cos A sin A) / (cos A sin A)]
=1+cscAsecA

Answer 15

for example the first fraction at the left side will become:
\[\Large \frac{{\tan A}}{{1 - \cot A}} = \frac{{\tan A}}{{1 - \frac{1}{{\tan A}}}} = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}}\]

Answer 16

second one

(cotA)^2/cotA - 1

Answer 17

right?

Answer 18

the second fraction will become:
\[\Large \begin{gathered}
\frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\
= - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\
\end{gathered} \]

Answer 19

Okay!
What next?

Answer 20

we have to add those fractions together, like below:
\[\Large \begin{gathered}
\frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\
\end{gathered} \]

Answer 21

yep that's what i got

Answer 22

next we have this factorization:
\[\large {\left( {\tan A} \right)^3} - 1 = \left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}\]

Answer 23

you have subtracted the 2 fraction ... why?

Answer 24

http://prntscr.com/7x8oyo

shouldnt the sign by +?

Answer 25

@Michele_Laino

Answer 26

I have not subtracted the 2 fractions, since the second fraction can be rewritten as follows:
\[\Large \begin{gathered}
\frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\
= - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\
\end{gathered} \]

Answer 27

ok!
what next?

Answer 28

using the factorization above, I can write:
\[\large \begin{gathered}
\frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} \hfill \\
\hfill \\
\end{gathered} \]

Answer 29

ok!

Answer 30

next we have:
\[\large \begin{gathered}
\frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\
\end{gathered} \]

Answer 31

i did not get the last time up there

Answer 32

now, we can simplify as follows:
\[\Large \begin{gathered}
\frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}}\frac{{\cos A}}{{\sin A}} = \hfill \\
\hfill \\
= \frac{1}{{\sin A\cos A}} = \sec A\csc A \hfill \\
\end{gathered} \]

Answer 33

and then:
\[\Large \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 = \sec A\csc A + 1\]

Answer 34

I didnt get a thing ......

Answer 35

ok! I redo my procedure

Answer 36

if we make this substitution:
\[\cot A = \frac{1}{{\tan A}}\]

Answer 37

we can write this:
\[\frac{{\tan A}}{{1 - \cot A}} = \frac{{\tan A}}{{1 - \frac{1}{{\tan A}}}} = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}}\]

Answer 38

and:
\[\begin{gathered}
\frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\
= - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\
\end{gathered} \]

Answer 39

am I right?

Answer 40

yes.

Answer 41

now I develop your original expression as follows:
\[\begin{gathered}
\frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\end{gathered} \]

Answer 42

ok?

Answer 43

are you sure?

Answer 44

yes..

Answer 45

ok! Now the common denominator is:
\[{\tan A\left( {\tan A - 1} \right)}\]

Answer 46

yes

Answer 47

so, I can write this:
\[\begin{gathered}
\frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\hfill \\
\end{gathered} \]

Answer 48

where, at the last step I have used my factorization:
\[{\left( {\tan A} \right)^3} - 1 = \left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}\]

Answer 49

ok

Answer 50

now I can cancel out
tanA-1 from the numerator and denominator of the last fraction

Answer 51

yup

Answer 52

so I get:
\[\begin{gathered}
\frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\
\end{gathered} \]

Answer 53

the last step ..me no get...

Answer 54

here is the last step:
\[\begin{gathered}
\frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1 + \tan A}}{{\tan A}} = \hfill \\
\hfill \\
= \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\} + \tan A}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\
\end{gathered} \]

Answer 55

shouldnt both the tan be cancelled?

Answer 56

not exactly, look at this:
\[\begin{gathered}
\frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1 + \tan A}}{{\tan A}} = \hfill \\
\hfill \\
= \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\} + \tan A}}{{\tan A}} = \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\}}}{{\tan A}} + \frac{{\tan A}}{{\tan A}} \hfill \\
\hfill \\
= \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\
\end{gathered} \]

Answer 57

okay!

Answer 58

ok! Now we use the substitution:
\[\tan A = \frac{{\sin A}}{{\cos A}}\]

Answer 59

ok

Answer 60

so we can write this:
\[\begin{gathered}
\frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \hfill \\
\hfill \\
= \frac{{\frac{1}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} = \frac{1}{{\sin A\cos A}} \hfill \\
\end{gathered} \]

Answer 61

\[\large \begin{gathered}
\frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \hfill \\
\hfill \\
= \frac{{\frac{1}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} = \frac{1}{{\sin A\cos A}} \hfill \\
\end{gathered} \]

Answer 62

THANKS!

Answer 63

:)