Answer :
Sure! Let's prove that:
[tex]\[ \frac{\tan x - \sin x}{\sin^3 x} = \frac{\sec x}{1 + \cos x} \][/tex]
Step 1: Simplify the Left-Hand Side (LHS)
Starting with the LHS:
[tex]\[ \frac{\tan x - \sin x}{\sin^3 x} \][/tex]
Recall the identity for \(\tan x\):
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
Substituting this into the LHS, we get:
[tex]\[ \frac{\frac{\sin x}{\cos x} - \sin x}{\sin^3 x} \][/tex]
Combine the terms in the numerator over a common denominator:
[tex]\[ \frac{\frac{\sin x - \sin x \cos x}{\cos x}}{\sin^3 x} \][/tex]
This can be written as:
[tex]\[ \frac{\sin x (1 - \cos x)}{\cos x \sin^3 x} \][/tex]
Simplify the fraction by canceling one \(\sin x\) in the numerator and denominator:
[tex]\[ \frac{1 - \cos x}{\cos x \sin^2 x} \][/tex]
Step 2: Simplify the Right-Hand Side (RHS)
Starting with the RHS:
[tex]\[ \frac{\sec x}{1 + \cos x} \][/tex]
Recall the identity for \(\sec x\):
[tex]\[ \sec x = \frac{1}{\cos x} \][/tex]
Substituting this into the RHS, we get:
[tex]\[ \frac{\frac{1}{\cos x}}{1 + \cos x} \][/tex]
This can be written as:
[tex]\[ \frac{1}{\cos x (1 + \cos x)} \][/tex]
Step 3: Compare the LHS and RHS
Now, we have the simplified forms of the LHS and the RHS:
[tex]\[ \text{LHS} = \frac{1 - \cos x}{\cos x \sin^2 x} \][/tex]
[tex]\[ \text{RHS} = \frac{1}{\cos x (1 + \cos x)} \][/tex]
To compare them, observe that the LHS can be written as:
[tex]\[ \text{LHS} = \frac{1 - \cos x}{\cos x \sin^2 x} \][/tex]
and the RHS is:
[tex]\[ \text{RHS} = \frac{1}{\cos x (1 + \cos x)} \][/tex]
For these to be equal, we would need:
[tex]\[ \frac{1 - \cos x}{\sin^2 x} = \frac{1}{1 + \cos x} \][/tex]
Recognize the following Pythagorean identity:
[tex]\[ \sin^2 x = 1 - \cos^2 x \][/tex]
Divide both sides of this identity by \(1 + \cos x\):
[tex]\[ \frac{\sin^2 x}{1 + \cos x} = \frac{1 - \cos^2 x}{1 + \cos x} = 1 - \cos x \][/tex]
Thus,
[tex]\[ 1 - \cos x = \frac{\sin^2 x}{1 + \cos x} \][/tex]
So,
[tex]\[ \frac{1 - \cos x}{\sin^2 x} = \frac{1}{1 + \cos x} \][/tex]
Therefore,
[tex]\[ \frac{\tan x - \sin x}{\sin^3 x} = \frac{\sec x}{1 + \cos x} \][/tex]
We have shown that the expression on the LHS is indeed equal to the expression on the RHS, proving the given identity.
[tex]\[ \frac{\tan x - \sin x}{\sin^3 x} = \frac{\sec x}{1 + \cos x} \][/tex]
Step 1: Simplify the Left-Hand Side (LHS)
Starting with the LHS:
[tex]\[ \frac{\tan x - \sin x}{\sin^3 x} \][/tex]
Recall the identity for \(\tan x\):
[tex]\[ \tan x = \frac{\sin x}{\cos x} \][/tex]
Substituting this into the LHS, we get:
[tex]\[ \frac{\frac{\sin x}{\cos x} - \sin x}{\sin^3 x} \][/tex]
Combine the terms in the numerator over a common denominator:
[tex]\[ \frac{\frac{\sin x - \sin x \cos x}{\cos x}}{\sin^3 x} \][/tex]
This can be written as:
[tex]\[ \frac{\sin x (1 - \cos x)}{\cos x \sin^3 x} \][/tex]
Simplify the fraction by canceling one \(\sin x\) in the numerator and denominator:
[tex]\[ \frac{1 - \cos x}{\cos x \sin^2 x} \][/tex]
Step 2: Simplify the Right-Hand Side (RHS)
Starting with the RHS:
[tex]\[ \frac{\sec x}{1 + \cos x} \][/tex]
Recall the identity for \(\sec x\):
[tex]\[ \sec x = \frac{1}{\cos x} \][/tex]
Substituting this into the RHS, we get:
[tex]\[ \frac{\frac{1}{\cos x}}{1 + \cos x} \][/tex]
This can be written as:
[tex]\[ \frac{1}{\cos x (1 + \cos x)} \][/tex]
Step 3: Compare the LHS and RHS
Now, we have the simplified forms of the LHS and the RHS:
[tex]\[ \text{LHS} = \frac{1 - \cos x}{\cos x \sin^2 x} \][/tex]
[tex]\[ \text{RHS} = \frac{1}{\cos x (1 + \cos x)} \][/tex]
To compare them, observe that the LHS can be written as:
[tex]\[ \text{LHS} = \frac{1 - \cos x}{\cos x \sin^2 x} \][/tex]
and the RHS is:
[tex]\[ \text{RHS} = \frac{1}{\cos x (1 + \cos x)} \][/tex]
For these to be equal, we would need:
[tex]\[ \frac{1 - \cos x}{\sin^2 x} = \frac{1}{1 + \cos x} \][/tex]
Recognize the following Pythagorean identity:
[tex]\[ \sin^2 x = 1 - \cos^2 x \][/tex]
Divide both sides of this identity by \(1 + \cos x\):
[tex]\[ \frac{\sin^2 x}{1 + \cos x} = \frac{1 - \cos^2 x}{1 + \cos x} = 1 - \cos x \][/tex]
Thus,
[tex]\[ 1 - \cos x = \frac{\sin^2 x}{1 + \cos x} \][/tex]
So,
[tex]\[ \frac{1 - \cos x}{\sin^2 x} = \frac{1}{1 + \cos x} \][/tex]
Therefore,
[tex]\[ \frac{\tan x - \sin x}{\sin^3 x} = \frac{\sec x}{1 + \cos x} \][/tex]
We have shown that the expression on the LHS is indeed equal to the expression on the RHS, proving the given identity.
