Question

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

Answer 1

@Michele_Laino

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

sec^2(A)

Answer 8

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

Answer 9

didn't get the last step

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

yes!

Answer 14

now continue to simplify

Answer 15

confused ;-;

Answer 16

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