Question

Prove the given trigonometric expression
${\left( {1 + \tan A\tan B} \right)^2} + {\left( {\tan A - \tan B} \right)^2} = {\sec ^2}A{\sec ^2}B$

Answer 1

Here we have given a trigonometric expression in which we have to prove LHS = RHS. First we will start from LHS and as given in LHS square of terms so first we open with formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and then using trigonometric formula like ${\sec ^2}x = 1 + {\tan ^2}x$ we will get RHS.

Complete step-by-step solution:
We have given,
To prove that
${\left( {1 + \tan A\tan B} \right)^2} + {\left( {\tan A - \tan B} \right)^2} = {\sec ^2}A{\sec ^2}B$
Taking LHS
${\left( {1 + \tan A\tan B} \right)^2} + {\left( {\tan A - \tan B} \right)^2} = {\sec ^2}A{\sec ^2}B$
First of all we will open the square bracket using the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
We get,
$1 + {\tan ^2}A{\tan ^2}B + 2\tan A\tan B + {\tan ^2}A + {\tan ^2}B - 2\tan A\tan B$
After cancel out we get
$1 + {\tan ^2}A{\tan ^2}B + {\tan ^2}A + {\tan ^2}B$
Now on taking ${\tan ^2}B$ common we get
$1 + {\tan ^2}A + {\tan ^2}B\left( {1 + {{\tan }^2}A} \right)$
Now on taking $\left( {1 + {{\tan }^2}A} \right)$ common we get
$\left( {1 + {{\tan }^2}A} \right)\left( {1 + {{\tan }^2}B} \right)$
Now we know that
$\left( {{{\sec }^2}A = 1 + {{\tan }^2}A} \right)\left( {{{\sec }^2}B = 1 + {{\tan }^2}B} \right)$ using these formula we can write
${\sec ^2}A{\sec ^2}B$
Which is equal to RHS.
Hence proved LHS = RHS

Note: Whenever we get this type of question the key concept of solving is we have to first think in our mind that either we should start from LHS side or RHS side that means we have to choose easy side and we have to remember formulae like $\left( {{{\sec }^2}A = 1 + {{\tan }^2}A} \right), \left( {{{\sec }^2}B = 1 + {{\tan }^2}B} \right)$ these formula help in solving this type of question.