AaronAndyson
  • AaronAndyson
(1+tan^2(A) times cot(A )all over cosec^2(A) = tan(A)
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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AaronAndyson
  • AaronAndyson
@Michele_Laino
Michele_Laino
  • Michele_Laino
here, 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
  • AaronAndyson
The question is \[(1 + \tan^2(A).\cot(A) \over cosec^2(A) \] is equal to tan(A)

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Michele_Laino
  • Michele_Laino
yes! it is an identity, so we have to show that the left side is equal to the right side
AaronAndyson
  • AaronAndyson
so sec^2(A).cot(A) over cosec^2(A)
Michele_Laino
  • Michele_Laino
sorry I have made an error, here are your right steps: \[\Large \begin{gathered} \frac{{\left( {1 + {{\left( {\tan A} \right)}^2}} \right)\cot A}}{{{{\left( {\csc A} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right)\frac{{\cos A}}{{\sin A}}}}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = ... \hfill \\ \end{gathered} \]
AaronAndyson
  • AaronAndyson
sec^2(A)
Michele_Laino
  • Michele_Laino
hint: 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
  • AaronAndyson
didn't get the last step
Michele_Laino
  • Michele_Laino
we 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
  • Michele_Laino
furthermore, 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
  • AaronAndyson
1/cos^2(A) . cos(A)/sin(A) . (sin^2(A))
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
now continue to simplify
AaronAndyson
  • AaronAndyson
confused ;-;
Michele_Laino
  • Michele_Laino
\[\Large |dw:1438936725328:dw|\frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = \frac{{\sin A}}{{\cos A}}\]
Michele_Laino
  • Michele_Laino
\[\Large \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = \frac{{\sin A}}{{\cos A}}\]

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