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AaronAndyson
 one year ago
(1+tan^2(A) times cot(A )all over cosec^2(A) = tan(A)
AaronAndyson
 one year ago
(1+tan^2(A) times cot(A )all over cosec^2(A) = tan(A)

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0here, a possible way, is to replace tanA with sinA/cosA, and cosecA with 1/sinA, in so doing the left side becomes: \[\Large \begin{gathered} \frac{{{{\left( {1 + \tan A} \right)}^2}\cot A}}{{{{\left( {\csc A} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {1 + \frac{{\sin A}}{{\cos A}}} \right)}^2}\frac{{\cos A}}{{\sin A}}}}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = ... \hfill \\ \end{gathered} \]

AaronAndyson
 one year ago
Best ResponseYou've already chosen the best response.0The question is \[(1 + \tan^2(A).\cot(A) \over cosec^2(A) \] is equal to tan(A)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0yes! it is an identity, so we have to show that the left side is equal to the right side

AaronAndyson
 one year ago
Best ResponseYou've already chosen the best response.0so sec^2(A).cot(A) over cosec^2(A)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0sorry I have made an error, here are your right steps: \[\Large \begin{gathered} \frac{{\left( {1 + {{\left( {\tan A} \right)}^2}} \right)\cot A}}{{{{\left( {\csc A} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right)\frac{{\cos A}}{{\sin A}}}}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = ... \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0hint: here are more steps: \[\Large \begin{gathered} \frac{{\left( {1 + {{\left( {\tan A} \right)}^2}} \right)\cot A}}{{{{\left( {\csc A} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right)\frac{{\cos A}}{{\sin A}}}}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = ... \hfill \\ \end{gathered} \]

AaronAndyson
 one year ago
Best ResponseYou've already chosen the best response.0didn't get the last step

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0we have the last step because we can write this: \[\Large \left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right) = \frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0furthermore, we can write this: \[\Large \frac{1}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = \frac{1}{{\frac{1}{{{{\left( {\sin A} \right)}^2}}}}} = {\left( {\sin A} \right)^2}\]

AaronAndyson
 one year ago
Best ResponseYou've already chosen the best response.01/cos^2(A) . cos(A)/sin(A) . (sin^2(A))

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0now continue to simplify

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large dw:1438936725328:dw\frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = \frac{{\sin A}}{{\cos A}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = \frac{{\sin A}}{{\cos A}}\]
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