Prove the identities. $$ 1 \frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x} $$ $$ 2 \frac{\cos ^{2} x-\cos x-\sin ^{2} x}{2 \sin x \cdot \cos x+\sin x}=\frac{1}{ an x}-\frac{1}{\sin x} $$

Question

Prove the identities. $$ 1 \frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x} $$ $$ 2 \frac{\cos ^{2} x-\cos x-\sin ^{2} x}{2 \sin x \cdot \cos x+\sin x}=\frac{1}{	an x}-\frac{1}{\sin x} $$

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We wish to show that

(1)  cos x/(1 – sin x) = (1 + sin x)/cos x  
  (2)  [cos²x – cos x – sin²x]⁄[2 sin x cos x + sin x] = (1/tan x) – (1/sin x).

Below are the detailed proofs for each identity.

────────────────────────────  
Proof of Identity (1):

Start with the left-hand side:

LHS = cos x/(1 – sin x).

Multiply the numerator and the denominator by (1 + sin x) (this is a common “trick” to obtain a difference of squares in the denominator):

cos x/(1 – sin x) = [cos x(1 + sin x)]/[(1 – sin x)(1 + sin x)].

Since (1 – sin x)(1 + sin x) is a difference of squares, we have:

(1 – sin x)(1 + sin x) = 1 – sin²x.

But recall the Pythagorean identity:

sin²x + cos²x = 1  ⟹  1 – sin²x = cos²x.

Thus, the expression becomes:

[cos x(1 + sin x)]/cos²x = (1 + sin x)/cos x.

This is exactly the right-hand side. Therefore, identity (1) is proven.

────────────────────────────  
Proof of Identity (2):

We want to show that

[cos²x – cos x – sin²x]⁄[2 sin x cos x + sin x] = (1/tan x) – (1/sin x).

Step 1. Rewrite the Right-Hand Side (RHS) in a simpler form.

Recall that tan x = sin x/cos x, so
  1/tan x = cos x/sin x.
Thus,
  (1/tan x) – (1/sin x) = cos x/sin x – 1/sin x = (cos x – 1)/sin x.

So our goal is to show that

[cos²x – cos x – sin²x]⁄[2 sin x cos x + sin x] = (cos x – 1)/sin x.

Step 2. Simplify the Left-Hand Side (LHS).

Notice that the denominator can be factored by taking sin x common:
  2 sin x cos x + sin x = sin x(2 cos x + 1).

So the LHS becomes:
  [cos²x – cos x – sin²x] / [sin x (2 cos x + 1)].

Step 3. Multiply both sides of the desired equality by sin x(2 cos x + 1) to remove fractions. Then the identity we need to prove becomes:

cos²x – cos x – sin²x = (cos x – 1)(2 cos x + 1).

Step 4. Expand the right-hand side:

(cos x – 1)(2 cos x + 1)  
    = cos x(2 cos x + 1) – 1(2 cos x + 1)  
    = 2 cos²x + cos x – 2 cos x – 1  
    = 2 cos²x – cos x – 1.

Now our equation to prove is:
  cos²x – cos x – sin²x = 2 cos²x – cos x – 1.

Step 5. Rearranging the equation by subtracting the left-hand side from the right-hand side gives:

[2 cos²x – cos x – 1] – [cos²x – cos x – sin²x]  
    = 2 cos²x – cos x – 1 – cos²x + cos x + sin²x  
    = cos²x + sin²x – 1.

By the Pythagorean identity, cos²x + sin²x = 1, so:

cos²x + sin²x – 1 = 1 – 1 = 0.

Since the difference is 0, the equation holds true. Thus, the given identity (2) is verified.

────────────────────────────  
Conclusion:

Both identities have been algebraically verified:

1) cos x/(1 – sin x) = (1 + sin x)/cos x  
2) [cos²x – cos x – sin²x]⁄[2 sin x cos x + sin x] = (1/tan x) – (1/sin x).

These verifications make use of standard trigonometric identities and algebraic manipulations.
Both identities are proven as follows: 1. $$ \frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x} $$ 2. $$ \frac{\cos^2 x - \cos x - \sin^2 x}{2 \sin x \cos x + \sin x} = \frac{1}{	an x} - \frac{1}{\sin x} $$ These proofs use trigonometric identities and algebraic manipulations to show that each side of the equations are equal.

Prove the identities. \( 1 \frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x} \) \( 2 \frac{\cos ^{2} x-\cos x-\sin ^{2} x}{2 \sin x \cdot \cos x+\sin x}=\frac{1}{\tan x}-\frac{1}{\sin x} \)

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