If \[x+y={{45}^{\circ }},\] then prove that
\[\left( a \right)\left( 1+\tan x \right)\left( 1+\tan y \right)=2\]
\[\left( b \right)\left( \cot x-1 \right)\left( \cot y-1 \right)=2\]

Hint: We have \[\left( x+y \right)={{45}^{\circ }}.\] We first have to show \[\left( 1+\tan x \right)\left( 1+\tan y \right)=2.\] So, we will first expand the brackets and then using \[\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\] we will get our required value. For the second part, we will first open the bracket and then we will use \[\cot \left( x+y \right)=\dfrac{\cot x\cot y-1}{\cot x+\cot y}\] and we will use \[x+y={{45}^{\circ }}\] to get our required solution.

Complete step by step answer:
We are given that \[x+y={{45}^{\circ }}\] we will first have to prove that \[\left( 1+\tan x \right)\left( 1+\tan y \right)=2.\] So, we will start by considering the left-hand side. So, we have,
\[\left( 1+\tan x \right)\left( 1+\tan y \right)\]
Opening the brackets, we get,
\[\Rightarrow \left( 1+\tan x \right)\left( 1+\tan y \right)=1+\tan x\tan y+\tan x+\tan y......\left( i \right)\]
Now, we will use the trigonometric formula which is expressed as \[\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}.\]
Now, as we have \[x+y={{45}^{\circ }}\] so using this in the above equation, we get,
\[\tan {{45}^{\circ }}=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\]
As, \[\tan {{45}^{\circ }}=1\] so we have,
\[1=\dfrac{\tan x+\tan y}{1-\tan x\tan y}\]
On simplifying, we get,
\[\Rightarrow 1\times \left( 1-\tan x\tan y \right)=\tan x+\tan y\]
So,
\[\Rightarrow 1-\tan x\tan y=\tan x+\tan y\]
Shifting tan x tan y to the other side, we get,
\[\Rightarrow 1=\tan x+\tan y+\tan x\tan y......\left( ii \right)\]
Now using (i) and (ii), we get,
\[\Rightarrow \left( 1+\tan x \right)\left( 1+\tan y \right)=1+\tan x+\tan y+\tan x\tan y\left[ \text{using (ii)} \right]\]
\[\Rightarrow \left( 1+\tan x \right)\left( 1+\tan y \right)=1+1\]
\[\Rightarrow \left( 1+\tan x \right)\left( 1+\tan y \right)=2\]
So, we get,
\[\Rightarrow \left( 1+\tan x \right)\left( 1+\tan y \right)=2\]
Hence proved
(b) Secondly, we have to prove that
\[\left( \cot x-1 \right)\left( \cot y-1 \right)=2\]
We start again by considering the left-hand side, so we get,
\[\left( \cot x-1 \right)\left( \cot y-1 \right)\]
Opening the brackets, we get,
\[\Rightarrow \left( \cot x-1 \right)\left( \cot y-1 \right)=\cot x\cot y-\cot x-\cot y+1......\left( iii \right)\]
Now we will use a trigonometric formula which is as follows
\[\cot \left( x+y \right)=\dfrac{\cot x\cot y-1}{\cot x+\cot y}\]
Now, as \[x+y={{45}^{\circ }}\] so we get,
\[\Rightarrow \cot {{45}^{\circ }}=\dfrac{\cot x\cot y-1}{\cot x+\cot y}\]
As \[\cot {{45}^{\circ }}=1\] so we get,
\[\Rightarrow 1=\dfrac{\cot x\cot y-1}{\cot x+\cot y}\]
Cross multiplying, we get,
\[\Rightarrow \cot y+\cot x=\cot x\cot y-1\]
Simplifying, we get,
\[\Rightarrow \cot x\cot y-\cot y-\cot x=1......\left( iv \right)\]
Now, using (iii) and (iv), we get,
\[\Rightarrow \left( \cot x-1 \right)\left( \cot y-1 \right)=\cot x\cot y-\cot x-\cot y+1\]
\[\Rightarrow \left( \cot x-1 \right)\left( \cot y-1 \right)=1+1\left[ \text{using (iv)} \right]\]
\[\Rightarrow \left( \cot x-1 \right)\left( \cot y-1 \right)=2\]
Hence proved

Note:
Students should remember when we product the elements of two brackets, then we have to multiply each element of the first bracket with each element of the other bracket.
\[\left( 1+\tan x \right)\left( 1+\tan y \right)\ne 1\times 1+\tan x\tan y\]
But \[\left( 1+\tan x \right)\left( 1+\tan y \right)=\left( 1+\tan x \right)\left( 1 \right)+\left( 1+\tan x \right)\left( \tan y \right)\]
\[\Rightarrow \left( 1+\tan x \right)\left( 1+\tan y \right)=1+\tan x+\tan y+\tan x\tan y\]
Also, while simplifying, we keep a close watch on the sign of the elements.