AaronAndyson
  • AaronAndyson
(1+tanA.tanB)^2 + (tanA - tanB)^2 = sec(A)^2 sec(B)^2
Mathematics
chestercat
  • chestercat
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AaronAndyson
  • AaronAndyson
AaronAndyson
  • AaronAndyson
Michele_Laino
  • Michele_Laino
we can try to use these identities: \[\begin{gathered} \tan A = \frac{{\sin A}}{{\cos A}} \hfill \\ \hfill \\ \tan B = \frac{{\sin B}}{{\cos B}} \hfill \\ \end{gathered} \]

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AaronAndyson
  • AaronAndyson
????
Michele_Laino
  • Michele_Laino
we can rewrite left side as follows:
AaronAndyson
  • AaronAndyson
as???
Michele_Laino
  • Michele_Laino
\[\begin{gathered} {\left( {1 + \tan A\tan B} \right)^2} = {\left( {1 + \frac{{\sin A}}{{\cos A}}\frac{{\sin B}}{{\cos B}}} \right)^2} = \hfill \\ \hfill \\ = {\left( {\frac{{\cos A\cos B + \sin A\sin B}}{{\cos A\cos B}}} \right)^2} = \hfill \\ \hfill \\ = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
it the first term
Michele_Laino
  • Michele_Laino
the second term will be as follows:
AaronAndyson
  • AaronAndyson
?
Michele_Laino
  • Michele_Laino
\[\begin{gathered} \tan A - \tan B = \frac{{\sin A}}{{\cos A}} - \frac{{\sin B}}{{\cos B}} = \hfill \\ \hfill \\ = \frac{{\sin A\cos B - \cos A\sin B}}{{\cos A\cos B}} = \hfill \\ \hfill \\ = \frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}} \hfill \\ \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
ok!
Michele_Laino
  • Michele_Laino
I think that maybe there is a typo in your original equation
Michele_Laino
  • Michele_Laino
please check your identity
AaronAndyson
  • AaronAndyson
Yep.Fixed it.
Michele_Laino
  • Michele_Laino
ok! so we can continue, and we can write this:
Michele_Laino
  • Michele_Laino
\[{\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2}\]
AaronAndyson
  • AaronAndyson
ok!
Michele_Laino
  • Michele_Laino
then we can write this next step: \[\begin{gathered} {\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2} = \hfill \\ \hfill \\ = \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} + \frac{{{{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
ok!
Michele_Laino
  • Michele_Laino
therefore, we get: \[\begin{gathered} {\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2} = \hfill \\ \hfill \\ = \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} + \frac{{{{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2} + {{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{1}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
may you explain the second last step?
Michele_Laino
  • Michele_Laino
in general we have this identity: (cos x)^2+ (sin x)^2 =1 for every value of x
AaronAndyson
  • AaronAndyson
ok
Michele_Laino
  • Michele_Laino
now we use these identities: \[\begin{gathered} \sec A = \frac{1}{{\cos A}} \hfill \\ \hfill \\ \sec B = \frac{1}{{\cos B}} \hfill \\ \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
ok
Michele_Laino
  • Michele_Laino
so we can rewrite the right side as follows: \[\begin{gathered} {\left( {\sec A} \right)^2}{\left( {\sec B} \right)^2} = {\left( {\frac{1}{{\cos A}}} \right)^2}{\left( {\frac{1}{{\cos B}}} \right)^2} = \hfill \\ \hfill \\ = \frac{1}{{{{\left( {\cos A} \right)}^2}}}\frac{1}{{{{\left( {\cos B} \right)}^2}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
thanks
Michele_Laino
  • Michele_Laino
:)

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