AaronAndyson
  • AaronAndyson
Trigo help..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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AaronAndyson
  • AaronAndyson
\[\frac{ \tan (A) }{ 1 - \cot(A) } + \frac{ \cot (A) }{ 1 - \tan(A) } = \sec(A)cosec(A) + 1\]
AaronAndyson
  • AaronAndyson
@Michele_Laino
AaronAndyson
  • AaronAndyson
@welshfella

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welshfella
  • welshfella
this one is tricky - i tried converting everything to sine and cosines but it only became more complicated.
welshfella
  • welshfella
it miggt help to convert all the cot A to tan A
welshfella
  • welshfella
- that doesn't help
AaronAndyson
  • AaronAndyson
@mathslover
AaronAndyson
  • AaronAndyson
@paki @IrishBoy123 @welshfella
Michele_Laino
  • Michele_Laino
please try this substitution: \[\cot A = \frac{1}{{\tan A}}\]
welshfella
  • welshfella
thats what I tried but didnt get very far - I'll have another try
AaronAndyson
  • AaronAndyson
It's getting complicated >.<
AaronAndyson
  • AaronAndyson
@Michele_Laino
welshfella
  • welshfella
i boiled it down to t^3 - 1 ------ t(t - 1) where t = tan A
imqwerty
  • imqwerty
(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
Michele_Laino
  • Michele_Laino
for example the first fraction at the left side will become: \[\Large \frac{{\tan A}}{{1 - \cot A}} = \frac{{\tan A}}{{1 - \frac{1}{{\tan A}}}} = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}}\]
AaronAndyson
  • AaronAndyson
second one (cotA)^2/cotA - 1
AaronAndyson
  • AaronAndyson
right?
Michele_Laino
  • Michele_Laino
the second fraction will become: \[\Large \begin{gathered} \frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\ = - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
Okay! What next?
Michele_Laino
  • Michele_Laino
we have to add those fractions together, like below: \[\Large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \end{gathered} \]
welshfella
  • welshfella
yep that's what i got
Michele_Laino
  • Michele_Laino
next we have this factorization: \[\large {\left( {\tan A} \right)^3} - 1 = \left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}\]
AaronAndyson
  • AaronAndyson
you have subtracted the 2 fraction ... why?
AaronAndyson
  • AaronAndyson
http://prntscr.com/7x8oyo shouldnt the sign by +?
AaronAndyson
  • AaronAndyson
@Michele_Laino
Michele_Laino
  • Michele_Laino
I have not subtracted the 2 fractions, since the second fraction can be rewritten as follows: \[\Large \begin{gathered} \frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\ = - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
ok! what next?
Michele_Laino
  • Michele_Laino
using the factorization above, I can write: \[\large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} \hfill \\ \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
ok!
Michele_Laino
  • Michele_Laino
next we have: \[\large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
i did not get the last time up there
Michele_Laino
  • Michele_Laino
now, we can simplify as follows: \[\Large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}}\frac{{\cos A}}{{\sin A}} = \hfill \\ \hfill \\ = \frac{1}{{\sin A\cos A}} = \sec A\csc A \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
and then: \[\Large \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 = \sec A\csc A + 1\]
AaronAndyson
  • AaronAndyson
I didnt get a thing ......
Michele_Laino
  • Michele_Laino
ok! I redo my procedure
Michele_Laino
  • Michele_Laino
if we make this substitution: \[\cot A = \frac{1}{{\tan A}}\]
Michele_Laino
  • Michele_Laino
we can write this: \[\frac{{\tan A}}{{1 - \cot A}} = \frac{{\tan A}}{{1 - \frac{1}{{\tan A}}}} = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}}\]
Michele_Laino
  • Michele_Laino
and: \[\begin{gathered} \frac{{\cot A}}{{1 - \tan A}} = \frac{{\frac{1}{{\tan A}}}}{{1 - \tan A}} = \frac{1}{{\tan A\left( {1 - \tan A} \right)}} = \hfill \\ = - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
am I right?
AaronAndyson
  • AaronAndyson
yes.
Michele_Laino
  • Michele_Laino
now I develop your original expression as follows: \[\begin{gathered} \frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
ok?
Michele_Laino
  • Michele_Laino
are you sure?
AaronAndyson
  • AaronAndyson
yes..
Michele_Laino
  • Michele_Laino
ok! Now the common denominator is: \[{\tan A\left( {\tan A - 1} \right)}\]
AaronAndyson
  • AaronAndyson
yes
Michele_Laino
  • Michele_Laino
so, I can write this: \[\begin{gathered} \frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
where, at the last step I have used my factorization: \[{\left( {\tan A} \right)^3} - 1 = \left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}\]
AaronAndyson
  • AaronAndyson
ok
Michele_Laino
  • Michele_Laino
now I can cancel out tanA-1 from the numerator and denominator of the last fraction
AaronAndyson
  • AaronAndyson
yup
Michele_Laino
  • Michele_Laino
so I get: \[\begin{gathered} \frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2}}}{{\tan A - 1}} - \frac{1}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^3} - 1}}{{\tan A\left( {\tan A - 1} \right)}} = \frac{{\left( {\tan A - 1} \right)\left\{ {{{\left( {\tan A} \right)}^2} + \tan A + 1} \right\}}}{{\tan A\left( {\tan A - 1} \right)}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
the last step ..me no get...
Michele_Laino
  • Michele_Laino
here is the last step: \[\begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1 + \tan A}}{{\tan A}} = \hfill \\ \hfill \\ = \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\} + \tan A}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
shouldnt both the tan be cancelled?
Michele_Laino
  • Michele_Laino
not exactly, look at this: \[\begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + \tan A + 1}}{{\tan A}} = \frac{{{{\left( {\tan A} \right)}^2} + 1 + \tan A}}{{\tan A}} = \hfill \\ \hfill \\ = \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\} + \tan A}}{{\tan A}} = \frac{{\left\{ {{{\left( {\tan A} \right)}^2} + 1} \right\}}}{{\tan A}} + \frac{{\tan A}}{{\tan A}} \hfill \\ \hfill \\ = \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} + 1 \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
okay!
Michele_Laino
  • Michele_Laino
ok! Now we use the substitution: \[\tan A = \frac{{\sin A}}{{\cos A}}\]
AaronAndyson
  • AaronAndyson
ok
Michele_Laino
  • Michele_Laino
so we can write this: \[\begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} = \frac{1}{{\sin A\cos A}} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
\[\large \begin{gathered} \frac{{{{\left( {\tan A} \right)}^2} + 1}}{{\tan A}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} + 1}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{{\frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \hfill \\ \hfill \\ = \frac{{\frac{1}{{{{\left( {\cos A} \right)}^2}}}}}{{\frac{{\sin A}}{{\cos A}}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} = \frac{1}{{\sin A\cos A}} \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
THANKS!
Michele_Laino
  • Michele_Laino
:)

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