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AaronAndyson

  • one year ago

(1+tan^2(A) times cot(A )all over cosec^2(A) = tan(A)

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  1. AaronAndyson
    • one year ago
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    @Michele_Laino

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

  3. AaronAndyson
    • one year ago
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    The question is \[(1 + \tan^2(A).\cot(A) \over cosec^2(A) \] is equal to tan(A)

  4. Michele_Laino
    • one year ago
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    yes! it is an identity, so we have to show that the left side is equal to the right side

  5. AaronAndyson
    • one year ago
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    so sec^2(A).cot(A) over cosec^2(A)

  6. Michele_Laino
    • one year ago
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    sorry I have made an error, here are your right steps: \[\Large \begin{gathered} \frac{{\left( {1 + {{\left( {\tan A} \right)}^2}} \right)\cot A}}{{{{\left( {\csc A} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right)\frac{{\cos A}}{{\sin A}}}}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = ... \hfill \\ \end{gathered} \]

  7. AaronAndyson
    • one year ago
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    sec^2(A)

  8. Michele_Laino
    • one year ago
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    hint: here are more steps: \[\Large \begin{gathered} \frac{{\left( {1 + {{\left( {\tan A} \right)}^2}} \right)\cot A}}{{{{\left( {\csc A} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right)\frac{{\cos A}}{{\sin A}}}}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = ... \hfill \\ \end{gathered} \]

  9. AaronAndyson
    • one year ago
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    didn't get the last step

  10. Michele_Laino
    • one year ago
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    we have the last step because we can write this: \[\Large \left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right) = \frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}\]

  11. Michele_Laino
    • one year ago
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    furthermore, we can write this: \[\Large \frac{1}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = \frac{1}{{\frac{1}{{{{\left( {\sin A} \right)}^2}}}}} = {\left( {\sin A} \right)^2}\]

  12. AaronAndyson
    • one year ago
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    1/cos^2(A) . cos(A)/sin(A) . (sin^2(A))

  13. Michele_Laino
    • one year ago
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    yes!

  14. Michele_Laino
    • one year ago
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    now continue to simplify

  15. AaronAndyson
    • one year ago
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    confused ;-;

  16. Michele_Laino
    • one year ago
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    \[\Large |dw:1438936725328:dw|\frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = \frac{{\sin A}}{{\cos A}}\]

  17. Michele_Laino
    • one year ago
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    \[\Large \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = \frac{{\sin A}}{{\cos A}}\]

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