Prove the following identities. $$ \begin{array}{l}1 \frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x} \ 2 \quad \frac{\cos ^{2} x-\cos x-\sin ^{2} x}{2 \sin x \cdot \cos x+\sin x}=\frac{1}{ an x}-\frac{1}{\sin x}\end{array} $$

Question

Prove the following identities. $$ \begin{array}{l}1 \frac{\cos x}{1-\sin x}=\frac{1+\sin x}{\cos x} \ 2 \quad \frac{\cos ^{2} x-\cos x-\sin ^{2} x}{2 \sin x \cdot \cos x+\sin x}=\frac{1}{	an x}-\frac{1}{\sin x}\end{array} $$

UpStudy AI · Accepted Answer

Sure! Let's prove each identity step by step.

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### **1. Prove that**
$$
\frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x}
$$

**Proof:**

Start with the left-hand side (LHS):
$$
\frac{\cos x}{1 - \sin x}
$$

To simplify, multiply the numerator and the denominator by the conjugate of the denominator, which is $$1 + \sin x$$:
$$
\frac{\cos x}{1 - \sin x} \times \frac{1 + \sin x}{1 + \sin x} = \frac{\cos x (1 + \sin x)}{(1 - \sin x)(1 + \sin x)}
$$

Simplify the denominator using the identity $$a^2 - b^2 = (a - b)(a + b)$$:
$$
(1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x
$$

Now, the expression becomes:
$$
\frac{\cos x (1 + \sin x)}{\cos^2 x} = \frac{1 + \sin x}{\cos x}
$$

Thus,
$$
\frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x}
$$
Q.E.D.

---

### **2. Prove that**
$$
\frac{\cos^{2} x - \cos x - \sin^{2} x}{2 \sin x \cos x + \sin x} = \frac{1}{\tan x} - \frac{1}{\sin x}
$$

**Proof:**

Let's start by simplifying the left-hand side (LHS):
$$
\frac{\cos^{2} x - \cos x - \sin^{2} x}{2 \sin x \cos x + \sin x}
$$

**Step 1:** Simplify the numerator.

Recall the Pythagorean identity:
$$
\cos^{2} x + \sin^{2} x = 1 \quad \Rightarrow \quad \sin^{2} x = 1 - \cos^{2} x
$$
Substitute $$\sin^{2} x$$ in the numerator:
$$
\cos^{2} x - \cos x - (1 - \cos^{2} x) = \cos^{2} x - \cos x - 1 + \cos^{2} x = 2\cos^{2} x - \cos x - 1
$$

**Step 2:** Factor the numerator.

We need to factor $$2\cos^{2} x - \cos x - 1$$. Let’s find two numbers that multiply to $$-2$$ (from $$2 \times -1$$) and add to $$-1$$:
$$
2\cos^{2} x - 2\cos x + \cos x - 1 = (2\cos^{2} x - 2\cos x) + (\cos x - 1) = 2\cos x(\cos x - 1) + 1(\cos x - 1) = (2\cos x + 1)(\cos x - 1)
$$
So, the numerator factors as:
$$
(2\cos x + 1)(\cos x - 1)
$$

**Step 3:** Simplify the denominator.

Factor out $$\sin x$$:
$$
2 \sin x \cos x + \sin x = \sin x (2\cos x + 1)
$$

**Step 4:** Combine numerator and denominator.
$$
\frac{(2\cos x + 1)(\cos x - 1)}{\sin x (2\cos x + 1)} = \frac{\cos x - 1}{\sin x}
$$
Here, $$2\cos x + 1$$ cancels out from the numerator and denominator.

**Step 5:** Simplify the right-hand side (RHS).

Compute RHS:
$$
\frac{1}{\tan x} - \frac{1}{\sin x} = \frac{\cos x}{\sin x} - \frac{1}{\sin x} = \frac{\cos x - 1}{\sin x}
$$

Thus, both LHS and RHS simplify to $$\frac{\cos x - 1}{\sin x}$$.

Therefore,
$$
\frac{\cos^{2} x - \cos x - \sin^{2} x}{2 \sin x \cos x + \sin x} = \frac{1}{\tan x} - \frac{1}{\sin x}
$$
Q.E.D.

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Both identities have been successfully proven!
Both identities are proven as follows: 1. $$ \frac{\cos x}{1 - \sin x} = \frac{1 + \sin x}{\cos x} $$ 2. $$ \frac{\cos^{2} x - \cos x - \sin^{2} x}{2 \sin x \cos x + \sin x} = \frac{1}{\tan x} - \frac{1}{\sin x} $$ These proofs show that both equations hold true for all values of $$ x $$ where the expressions are defined.

Prove the following identities.

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