Answer :
To solve the given equation [tex]\(\frac{(1+\tan x)^2}{\sec x} = \sec x + 2 \sin x\)[/tex], we will simplify both sides using trigonometric identities.
### Step 1: Simplify the left-hand side (LHS)
We begin with:
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} \][/tex]
First, recall the trigonometric identities:
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
[tex]\[ \sec x = \frac{1}{\cos x} \][/tex]
Using these identities, we can rewrite [tex]\(1 + \tan x\)[/tex] as:
[tex]\[ 1 + \tan x = 1 + \frac{\sin x}{\cos x} = \frac{\cos x + \sin x}{\cos x} \][/tex]
Then, square it:
[tex]\[ (1 + \tan x)^2 = \left( \frac{\cos x + \sin x}{\cos x} \right)^2 = \frac{(\cos x + \sin x)^2}{\cos^2 x} \][/tex]
Now, dividing by [tex]\(\sec x\)[/tex]:
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} = \frac{(\cos x + \sin x)^2}{\cos^2 x} \cdot \cos x = \frac{(\cos x + \sin x)^2}{\cos x} \][/tex]
### Step 2: Expand and simplify
Expand [tex]\((\cos x + \sin x)^2\)[/tex]:
[tex]\[ (\cos x + \sin x)^2 = \cos^2 x + 2 \cos x \sin x + \sin^2 x \][/tex]
Using the Pythagorean identity [tex]\(\cos^2 x + \sin^2 x = 1\)[/tex], we have:
[tex]\[ \cos^2 x + \sin^2 x = 1 \][/tex]
Substitute back into our expression:
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} = \frac{1 + 2 \cos x \sin x}{\cos x} \][/tex]
### Step 3: Simplify further
Note that:
[tex]\[ 2 \cos x \sin x = \sin(2x) \][/tex]
However, it may not directly help us here. Instead, we proceed by simply substituting and separating terms in the numerator:
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} = \frac{1}{\cos x} + \frac{2 \cos x \sin x}{\cos x} \][/tex]
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} = \sec x + 2 \sin x \][/tex]
### Step 4: Compare with the right-hand side (RHS)
We already have:
[tex]\[ \sec x + 2 \sin x \][/tex]
So our simplified LHS becomes:
[tex]\[ \sec x + 2 \sin x \][/tex]
Which exactly matches the RHS:
[tex]\[ \sec x + 2 \sin x \][/tex]
Thus, we have verified that:
[tex]\[ \frac{(1+\tan x)^2}{\sec x} = \sec x + 2 \sin x \][/tex]
Hence, the given equation is an identity and is satisfied for all [tex]\(x\)[/tex] where the trigonometric functions are defined.
### Step 1: Simplify the left-hand side (LHS)
We begin with:
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} \][/tex]
First, recall the trigonometric identities:
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
[tex]\[ \sec x = \frac{1}{\cos x} \][/tex]
Using these identities, we can rewrite [tex]\(1 + \tan x\)[/tex] as:
[tex]\[ 1 + \tan x = 1 + \frac{\sin x}{\cos x} = \frac{\cos x + \sin x}{\cos x} \][/tex]
Then, square it:
[tex]\[ (1 + \tan x)^2 = \left( \frac{\cos x + \sin x}{\cos x} \right)^2 = \frac{(\cos x + \sin x)^2}{\cos^2 x} \][/tex]
Now, dividing by [tex]\(\sec x\)[/tex]:
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} = \frac{(\cos x + \sin x)^2}{\cos^2 x} \cdot \cos x = \frac{(\cos x + \sin x)^2}{\cos x} \][/tex]
### Step 2: Expand and simplify
Expand [tex]\((\cos x + \sin x)^2\)[/tex]:
[tex]\[ (\cos x + \sin x)^2 = \cos^2 x + 2 \cos x \sin x + \sin^2 x \][/tex]
Using the Pythagorean identity [tex]\(\cos^2 x + \sin^2 x = 1\)[/tex], we have:
[tex]\[ \cos^2 x + \sin^2 x = 1 \][/tex]
Substitute back into our expression:
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} = \frac{1 + 2 \cos x \sin x}{\cos x} \][/tex]
### Step 3: Simplify further
Note that:
[tex]\[ 2 \cos x \sin x = \sin(2x) \][/tex]
However, it may not directly help us here. Instead, we proceed by simply substituting and separating terms in the numerator:
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} = \frac{1}{\cos x} + \frac{2 \cos x \sin x}{\cos x} \][/tex]
[tex]\[ \frac{(1 + \tan x)^2}{\sec x} = \sec x + 2 \sin x \][/tex]
### Step 4: Compare with the right-hand side (RHS)
We already have:
[tex]\[ \sec x + 2 \sin x \][/tex]
So our simplified LHS becomes:
[tex]\[ \sec x + 2 \sin x \][/tex]
Which exactly matches the RHS:
[tex]\[ \sec x + 2 \sin x \][/tex]
Thus, we have verified that:
[tex]\[ \frac{(1+\tan x)^2}{\sec x} = \sec x + 2 \sin x \][/tex]
Hence, the given equation is an identity and is satisfied for all [tex]\(x\)[/tex] where the trigonometric functions are defined.
