Question

Prove that, \[{{\left( 1+\tan A.\tan B \right)}^{2}}+{{\left( \tan A-\tan B \right)}^{2}}={{\sec }^{2}}A{{\sec }^{2}}B\].

Answer 1

Hint: The LHS and RHS of the given expression is \[{{\left( 1+\tan A.\tan B \right)}^{2}}+\left( \tan A-\tan B \right)\] and
\[{{\sec }^{2}}A{{\sec }^{2}}B\] respectively. Use the formulas, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] , and expand the LHS of the expression. After expanding the LHS, simplify it further. Now, the identity \[{{\sec }^{2}}x-{{\tan }^{2}}x=1\Rightarrow \sec x={{\tan }^{2}}x+1\] and solve it further.

Complete step-by-step solution:
According to the question, we are asked to prove \[{{\left( 1+\tan A.\tan B \right)}^{2}}+{{\left( \tan A-\tan B \right)}^{2}}={{\sec }^{2}}A{{\sec }^{2}}B\] .
In the LHS, we have
\[{{\left( 1+\tan A.\tan B \right)}^{2}}+{{\left( \tan A-\tan B \right)}^{2}}\] ………………………………….(1)
Similarly, in RHS we have
\[{{\sec }^{2}}A{{\sec }^{2}}B\] ……………………………………….(2)
To prove this, we have to make the LHS equal to RHS of the given expression.
Let us solve the LHS side first.
We know the formula, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] ………………………………………….(3)
Now, on replacing \[a\] by 1 and \[b\] by \[\tan A.\tan B\] in equation (3), we get
\[\Rightarrow {{\left( 1+\tan A.\tan B \right)}^{2}}={{1}^{2}}+{{\left( \tan A.\tan B \right)}^{2}}+2\left( 1 \right)\left( \tan A.\tan B \right)=1+{{\tan }^{2}}A.{{\tan }^{2}}B+2\tan A.\tan B\]…………………………………………(4)
 We also know the formula, \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] ………………………………………….(5)
Now, on replacing \[a\] by \[\tan A\] and \[b\] by \[\tan B\] in equation (5), we get
\[\Rightarrow {{\left( \tan A-\tan B \right)}^{2}}={{\left( \tan A \right)}^{2}}+{{\left( \tan B \right)}^{2}}-2\tan A.\tan B={{\tan }^{2}}A+{{\tan }^{2}}B-2\tan A.\tan B\] …………………………………………………..(6)
Now, from equation (1), equation (4), and equation (6), we get
\[={{\left( 1+\tan A.\tan B \right)}^{2}}+{{\left( \tan A-\tan B \right)}^{2}}\]
\[=1+{{\tan }^{2}}A.{{\tan }^{2}}B+2\tan A.\tan B+{{\tan }^{2}}A+{{\tan }^{2}}B-2\tan A.\tan B\] ………………………………….(7)
On arranging and modifying the above equation, we get
\[\begin{align}
  & =1+{{\tan }^{2}}A.{{\tan }^{2}}B+{{\tan }^{2}}A+{{\tan }^{2}}B \\
 & =1+{{\tan }^{2}}A+{{\tan }^{2}}A.{{\tan }^{2}}B+{{\tan }^{2}}B \\
\end{align}\]
\[=\left( 1+{{\tan }^{2}}A \right)+{{\tan }^{2}}B\left( 1+{{\tan }^{2}}A \right)\] ………………………………………….(8)
Now, on taking the term \[\left( 1+{{\tan }^{2}}A \right)\] as common from the whole, we get
\[=\left( 1+{{\tan }^{2}}A \right)\left( 1+{{\tan }^{2}}B \right)\] ………………………………………………(9)
We know the identity \[{{\sec }^{2}}x-{{\tan }^{2}}x=1\] ……………………………..(10)
Replacing \[x\] by \[A\] in equation (10), we get
 \[\Rightarrow {{\sec }^{2}}A-{{\tan }^{2}}A=1\]
\[\Rightarrow {{\sec }^{2}}A=1+{{\tan }^{2}}A\] ……………………………………(11)
Similarly, on replacing \[x\] by \[B\] in equation (10), we get
\[\Rightarrow {{\sec }^{2}}B-{{\tan }^{2}}B=1\]
\[\Rightarrow {{\sec }^{2}}B=1+{{\tan }^{2}}B\] ……………………………………(12)
Now, on replacing \[\left( 1+{{\tan }^{2}}A \right)\] by \[{{\sec }^{2}}A\] and \[\left( 1+{{\tan }^{2}}B \right)\] by \[{{\sec }^{2}}B\] in equation (9), we get
\[={{\sec }^{2}}A.{{\sec }^{2}}B\] ………………………………………….(13)
From equation (13), we have the LHS equal to \[{{\sec }^{2}}A.{{\sec }^{2}}B\] .
Similarly, from equation (2), we have the RHS equal to \[{{\sec }^{2}}A.{{\sec }^{2}}B\] .
Therefore, we can say that LHS = RHS.
Hence, proved.

Note: In this question, one might think to simplify the LHS of the given expression using the property, \[\tan x=\dfrac{\sin x}{\cos x}\] . Doing this may lead to complexity that can result in calculation mistakes. Therefore, avoid using the property, \[\tan x=\dfrac{\sin x}{\cos x}\] for the simplification of the given expression.