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

1. AaronAndyson

@Michele_Laino

2. 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: $\Large \begin{gathered} \frac{{{{\left( {1 + \tan A} \right)}^2}\cot A}}{{{{\left( {\csc A} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {1 + \frac{{\sin A}}{{\cos A}}} \right)}^2}\frac{{\cos A}}{{\sin A}}}}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = ... \hfill \\ \end{gathered}$

3. AaronAndyson

The question is $(1 + \tan^2(A).\cot(A) \over cosec^2(A)$ is equal to tan(A)

4. Michele_Laino

yes! it is an identity, so we have to show that the left side is equal to the right side

5. AaronAndyson

so sec^2(A).cot(A) over cosec^2(A)

6. Michele_Laino

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

7. AaronAndyson

sec^2(A)

8. Michele_Laino

hint: here are more steps: $\Large \begin{gathered} \frac{{\left( {1 + {{\left( {\tan A} \right)}^2}} \right)\cot A}}{{{{\left( {\csc A} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{\left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right)\frac{{\cos A}}{{\sin A}}}}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = \hfill \\ \hfill \\ = \frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = ... \hfill \\ \end{gathered}$

9. AaronAndyson

didn't get the last step

10. Michele_Laino

we have the last step because we can write this: $\Large \left( {1 + {{\left( {\frac{{\sin A}}{{\cos A}}} \right)}^2}} \right) = \frac{{{{\left( {\sin A} \right)}^2} + {{\left( {\cos A} \right)}^2}}}{{{{\left( {\cos A} \right)}^2}}}$

11. Michele_Laino

furthermore, we can write this: $\Large \frac{1}{{{{\left( {\frac{1}{{\sin A}}} \right)}^2}}} = \frac{1}{{\frac{1}{{{{\left( {\sin A} \right)}^2}}}}} = {\left( {\sin A} \right)^2}$

12. AaronAndyson

1/cos^2(A) . cos(A)/sin(A) . (sin^2(A))

13. Michele_Laino

yes!

14. Michele_Laino

now continue to simplify

15. AaronAndyson

confused ;-;

16. Michele_Laino

$\Large |dw:1438936725328:dw|\frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = \frac{{\sin A}}{{\cos A}}$

17. Michele_Laino

$\Large \frac{1}{{{{\left( {\cos A} \right)}^2}}} \cdot \frac{{\cos A}}{{\sin A}} \cdot {\left( {\sin A} \right)^2} = \frac{{\sin A}}{{\cos A}}$