Question

(1+tanA.tanB)^2 + (tanA - tanB)^2 = sec(A)^2 sec(B)^2

Answer 1

@welshfella

Answer 2

@Michele_Laino

Answer 3

we can try to use these identities:
\[\begin{gathered}
\tan A = \frac{{\sin A}}{{\cos A}} \hfill \\
\hfill \\
\tan B = \frac{{\sin B}}{{\cos B}} \hfill \\
\end{gathered} \]

Answer 4

????

Answer 5

we can rewrite left side as follows:

Answer 6

as???

Answer 7

\[\begin{gathered}
{\left( {1 + \tan A\tan B} \right)^2} = {\left( {1 + \frac{{\sin A}}{{\cos A}}\frac{{\sin B}}{{\cos B}}} \right)^2} = \hfill \\
\hfill \\
= {\left( {\frac{{\cos A\cos B + \sin A\sin B}}{{\cos A\cos B}}} \right)^2} = \hfill \\
\hfill \\
= {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} \hfill \\
\end{gathered} \]

Answer 8

it the first term

Answer 9

the second term will be as follows:

Answer 10

?

Answer 11

\[\begin{gathered}
\tan A - \tan B = \frac{{\sin A}}{{\cos A}} - \frac{{\sin B}}{{\cos B}} = \hfill \\
\hfill \\
= \frac{{\sin A\cos B - \cos A\sin B}}{{\cos A\cos B}} = \hfill \\
\hfill \\
= \frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}} \hfill \\
\hfill \\
\end{gathered} \]

Answer 12

ok!

Answer 13

I think that maybe there is a typo in your original equation

Answer 14

please check your identity

Answer 15

Yep.Fixed it.

Answer 16

ok! so we can continue, and we can write this:

Answer 17

\[{\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2}\]

Answer 18

ok!

Answer 19

then we can write this next step:
\[\begin{gathered}
{\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2} = \hfill \\
\hfill \\
= \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} + \frac{{{{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\
\end{gathered} \]

Answer 20

ok!

Answer 21

therefore, we get:
\[\begin{gathered}
{\text{left side}} = {\left( {\frac{{\cos \left( {A - B} \right)}}{{\cos A\cos B}}} \right)^2} + {\left( {\frac{{\left( {\sin A - B} \right)}}{{\cos A\cos B}}} \right)^2} = \hfill \\
\hfill \\
= \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} + \frac{{{{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} = \hfill \\
\hfill \\
= \frac{{{{\left\{ {\cos \left( {A - B} \right)} \right\}}^2} + {{\left\{ {\left( {\sin A - B} \right)} \right\}}^2}}}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} = \hfill \\
\hfill \\
= \frac{1}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\
\end{gathered} \]

Answer 22

may you explain the second last step?

Answer 23

in general we have this identity:

(cos x)^2+ (sin x)^2 =1
for every value of x

Answer 24

ok

Answer 25

now we use these identities:
\[\begin{gathered}
\sec A = \frac{1}{{\cos A}} \hfill \\
\hfill \\
\sec B = \frac{1}{{\cos B}} \hfill \\
\hfill \\
\end{gathered} \]

Answer 26

ok

Answer 27

so we can rewrite the right side as follows:
\[\begin{gathered}
{\left( {\sec A} \right)^2}{\left( {\sec B} \right)^2} = {\left( {\frac{1}{{\cos A}}} \right)^2}{\left( {\frac{1}{{\cos B}}} \right)^2} = \hfill \\
\hfill \\
= \frac{1}{{{{\left( {\cos A} \right)}^2}}}\frac{1}{{{{\left( {\cos B} \right)}^2}}} = \frac{1}{{{{\left( {\cos A} \right)}^2}{{\left( {\cos B} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]

Answer 28

thanks

Answer 29

:)