Question

Prove that secA(1-sinA)(secA+tanA)=1?

Answer 1

@Michele_Laino

Answer 2

here we have to use these identities:
\[\Large \sec A = \frac{1}{{\cos A}},\quad \tan A = \frac{{\sin A}}{{\cos A}}\]

Answer 3

hint:
we can write this step:
\[\Large \begin{gathered}
\sec A\left( {1 - \sin A} \right)\left( {\sec A + \tan A} \right) = \hfill \\
\hfill \\
= \frac{1}{{\cos A}}\left( {1 - \sin A} \right)\left( {\frac{1}{{\cos A}} + \frac{{\sin A}}{{\cos A}}} \right) = ...? \hfill \\
\end{gathered} \]
please, simplify

Answer 4

???

Answer 5

I have replaced secA wit 1/cosA and tanA with (sinA)/(cosA)

Answer 6

ok!

Answer 7

what next?>.<

Answer 8

hint:
\[\Large \begin{gathered}
\sec A\left( {1 - \sin A} \right)\left( {\sec A + \tan A} \right) = \hfill \\
\hfill \\
= \frac{1}{{\cos A}}\left( {1 - \sin A} \right)\left( {\frac{1}{{\cos A}} + \frac{{\sin A}}{{\cos A}}} \right) = \hfill \\
\hfill \\
= \frac{{1 - \sin A}}{{\cos A}} \cdot \frac{{1 + \sin A}}{{\cos A}} = ...? \hfill \\
\end{gathered} \]
now it is easy to write the next step

Answer 9

cross multiply?

Answer 10

you have to multiply those two fractions

Answer 11

??

Answer 12

hint:
\[\large \begin{gathered}
\sec A\left( {1 - \sin A} \right)\left( {\sec A + \tan A} \right) = \hfill \\
\hfill \\
= \frac{1}{{\cos A}}\left( {1 - \sin A} \right)\left( {\frac{1}{{\cos A}} + \frac{{\sin A}}{{\cos A}}} \right) = \hfill \\
\hfill \\
= \frac{{1 - \sin A}}{{\cos A}} \cdot \frac{{1 + \sin A}}{{\cos A}} = \hfill \\
\hfill \\
= \frac{{\left( {1 - \sin A} \right)\left( {1 + \sin A} \right)}}{{{{\left( {\cos A} \right)}^2}}} = ...? \hfill \\
\end{gathered} \]

Answer 13

1-sin^2/cos^2 is 1

Answer 14

that's right!

Answer 15

thanks help me more?

Answer 16

ok!