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

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

& =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{aligned}$$ $$=\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.