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AaronAndyson
 one year ago
(1+tanA.tanB)^2 + (tanA  tanB)^2 = sec(A)^2 sec(B)^2
AaronAndyson
 one year ago
(1+tanA.tanB)^2 + (tanA  tanB)^2 = sec(A)^2 sec(B)^2

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can try to use these identities: \[\begin{gathered} \tan A = \frac{{\sin A}}{{\cos A}} \hfill \\ \hfill \\ \tan B = \frac{{\sin B}}{{\cos B}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can rewrite left side as follows:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{gathered} {\left( {1 + \tan A\tan B} \right)^2} = {\left( {1 + \frac{{\sin A}}{{\cos A}}\frac{{\sin B}}{{\cos B}}} \right)^2} = \hfill \\ \hfill \\ = {\left( {\frac{{\cos A\cos B + \sin A\sin B}}{{\cos A\cos B}}} \right)^2} = \hfill \\ \hfill \\ = {\left( {\frac{{\cos \left( {A  B} \right)}}{{\cos A\cos B}}} \right)^2} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it the first term

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the second term will be as follows:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[\begin{gathered} \tan A  \tan B = \frac{{\sin A}}{{\cos A}}  \frac{{\sin B}}{{\cos B}} = \hfill \\ \hfill \\ = \frac{{\sin A\cos B  \cos A\sin B}}{{\cos A\cos B}} = \hfill \\ \hfill \\ = \frac{{\left( {\sin A  B} \right)}}{{\cos A\cos B}} \hfill \\ \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that maybe there is a typo in your original equation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please check your identity

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! so we can continue, and we can write this:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1\[{\text{left side}} = {\left( {\frac{{\cos \left( {A  B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A  B} \right)}}{{\cos A\cos B}}} \right)^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then we can write this next step: \[\begin{gathered} {\text{left side}} = {\left( {\frac{{\cos \left( {A  B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A  B} \right)}}{{\cos A\cos B}}} \right)^2} = \hfill \\ \hfill \\ = \frac{{{{\left\{ {\cos \left( {A  B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} + \frac{{{{\left\{ {\left( {\sin A  B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1therefore, 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
 one year ago
Best ResponseYou've already chosen the best response.0may you explain the second last step?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in general we have this identity: (cos x)^2+ (sin x)^2 =1 for every value of x

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we use these identities: \[\begin{gathered} \sec A = \frac{1}{{\cos A}} \hfill \\ \hfill \\ \sec B = \frac{1}{{\cos B}} \hfill \\ \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we can rewrite the right side as follows: \[\begin{gathered} {\left( {\sec A} \right)^2}{\left( {\sec B} \right)^2} = {\left( {\frac{1}{{\cos A}}} \right)^2}{\left( {\frac{1}{{\cos B}}} \right)^2} = \hfill \\ \hfill \\ = \frac{1}{{{{\left( {\cos A} \right)}^2}}}\frac{1}{{{{\left( {\cos B} \right)}^2}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered} \]
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