How do you prove $ {\left( {\sin x + \cos x} \right)^2} = 1 + 2\sin x\cos x $ ?

Hint: In this question we need to prove $ {\left( {\sin x + \cos x} \right)^2} = 1 + 2\sin x\cos x $ .
In order to prove we will consider the LHS and evaluate it using the formula of $ {\left( {a + b} \right)^2} $ . Then, we will apply the trigonometric formula and evaluate it to determine the required proof.

Complete step-by-step answer:
Here, we will prove $ {\left( {\sin x + \cos x} \right)^2} = 1 + 2\sin x\cos x $
Now let us consider the LHS,
LHS $ = {\left( {\sin x + \cos x} \right)^2} $
So, we can see that $ {\left( {\sin x + \cos x} \right)^2} $ is in the form of $ {\left( {a + b} \right)^2} $ .
Now, we know that $ {\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2} $
Here, $ a = \sin x $ and $ b = \cos x $
Thus, we will substitute the value of $ a $ and $ b $ in the formula, we have,
  $ {\left( {\sin x + \cos x} \right)^2} = {\sin ^2}x + 2\sin x\cos x + {\cos ^2}x $
Now, let us reorder the terms,
 $ {\left( {\sin x + \cos x} \right)^2} = {\sin ^2}x + {\cos ^2}x + 2\sin x\cos x $
From trigonometric identities we know that $ {\sin ^2}x + {\cos ^2}x = 1 $ .
Let us apply the value here, we have,
 $ {\left( {\sin x + \cos x} \right)^2} = 1 + 2\sin x\cos x $
Therefore, LHS=RHS
Hence proved.
So, the correct answer is “ $ {\left( {\sin x + \cos x} \right)^2} = 1 + 2\sin x\cos x $ ”.

Note: In this question it is important to note that whenever we come across these kinds of questions, we always start from the more complex side. This is because it is a lot easier to eliminate terms to make a complex function simple than to find ways to introduce terms to make a simple function complex. Take one step, watch one step.