1. Prove the following: $$ \begin{array}{l}\text { a) } \frac{2 \tan x-\sin 2 x}{2 \sin ^{2} x}=\tan x \ \text { b) } \tan ^{2} x+1=\sec ^{2} x \ \text { c) } \tan x+\frac{\cos x}{1+\sin x}=\sec x \ \text { d) } \frac{\tan x \sin x}{\tan x+\sin x}=\frac{\tan x-\sin x}{\operatorname{can} x \sin x} \ \text { e) } \frac{\tan x+\sec x}{\sec x\left(1+\frac{\tan x}{\operatorname{sen} x}\right)}=1 . \quad(20 \text { marks) }\end{array} $$

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Verify the identity by following steps:
- step0: Verify:
  $$\tan^{2}\left(x\right)+1=\sec^{2}\left(x\right)$$
- step1: Choose a side to work on:
  $$1+\tan^{2}\left(x\right)=\sec^{2}\left(x\right)$$
- step2: Choose the other side to work on:
  $$1+\tan^{2}\left(x\right)=\cos^{-2}\left(x\right)$$
- step3: Verify the identity:
  $$\textrm{false}$$
Determine whether the expression $$ \frac{\tan x+\sec x}{\sec x\left(1+\frac{\tan x}{\operatorname{sen} x}\right)}=1 $$ is always true.
Verify the identity by following steps:
- step0: Verify:
  $$\frac{\tan\left(x\right)+\sec\left(x\right)}{\sec\left(x\right)\left(1+\frac{\tan\left(x\right)}{\sin\left(x\right)}\right)}=1$$
- step1: Choose a side to work on:
  $$\frac{\cos\left(x\right)\sin\left(x\right)}{\cos\left(x\right)+1}+\frac{\cos\left(x\right)}{\cos\left(x\right)+1}=1$$
- step2: Verify the identity:
  $$\textrm{false}$$
Determine whether the expression $$ \tan x+\frac{\cos x}{1+\sin x}=\sec x $$ is always true.
Verify the identity by following steps:
- step0: Verify:
  $$\tan\left(x\right)+\frac{\cos\left(x\right)}{1+\sin\left(x\right)}=\sec\left(x\right)$$
- step1: Choose the other side to work on:
  $$\tan\left(x\right)+\frac{\cos\left(x\right)}{1+\sin\left(x\right)}=\frac{1}{\cos\left(x\right)}$$
- step2: Verify the identity:
  $$\textrm{false}$$
Determine whether the expression $$ \frac{2 \tan x-\sin 2 x}{2 \sin ^{2} x}=\tan x $$ is always true.
Verify the identity by following steps:
- step0: Verify:
  $$\frac{2\tan\left(x\right)-\sin\left(2x\right)}{2\sin^{2}\left(x\right)}=\tan\left(x\right)$$
- step1: Choose a side to work on:
  $$\tan\left(x\right)=\tan\left(x\right)$$
- step2: Verify the identity:
  $$\textrm{true}$$
Let's analyze each part of the problem step by step.

### a) Prove that 
$$
\frac{2 \tan x - \sin 2x}{2 \sin^2 x} = \tan x
$$

**Solution:**
1. Start with the left-hand side:
   $$
   \frac{2 \tan x - \sin 2x}{2 \sin^2 x}
   $$
2. Recall that $$\sin 2x = 2 \sin x \cos x$$, so substitute:
   $$
   = \frac{2 \tan x - 2 \sin x \cos x}{2 \sin^2 x}
   $$
3. Factor out the 2:
   $$
   = \frac{2(\tan x - \sin x \cos x)}{2 \sin^2 x} = \frac{\tan x - \sin x \cos x}{\sin^2 x}
   $$
4. Recall that $$\tan x = \frac{\sin x}{\cos x}$$:
   $$
   = \frac{\frac{\sin x}{\cos x} - \sin x \cos x}{\sin^2 x}
   $$
5. Combine the terms in the numerator:
   $$
   = \frac{\sin x \left(\frac{1}{\cos x} - \cos x\right)}{\sin^2 x} = \frac{\sin x \left(\frac{1 - \cos^2 x}{\cos x}\right)}{\sin^2 x}
   $$
6. Use the identity $$1 - \cos^2 x = \sin^2 x$$:
   $$
   = \frac{\sin x \cdot \frac{\sin^2 x}{\cos x}}{\sin^2 x} = \frac{\sin x}{\cos x} = \tan x
   $$
Thus, the statement is **true**.

### b) Prove that 
$$
\tan^2 x + 1 = \sec^2 x
$$

**Solution:**
This is a well-known trigonometric identity. The identity states that:
$$
\tan^2 x + 1 = \sec^2 x
$$
Thus, the statement is **true**.

### c) Prove that 
$$
\tan x + \frac{\cos x}{1 + \sin x} = \sec x
$$

**Solution:**
1. Start with the left-hand side:
   $$
   \tan x + \frac{\cos x}{1 + \sin x}
   $$
2. Recall that $$\tan x = \frac{\sin x}{\cos x}$$:
   $$
   = \frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x}
   $$
3. Find a common denominator:
   $$
   = \frac{\sin x(1 + \sin x) + \cos^2 x}{\cos x(1 + \sin x)}
   $$
4. Expand the numerator:
   $$
   = \frac{\sin x + \sin^2 x + \cos^2 x}{\cos x(1 + \sin x)}
   $$
5. Use the identity $$\sin^2 x + \cos^2 x = 1$$:
   $$
   = \frac{\sin x + 1}{\cos x(1 + \sin x)} = \frac{1 + \sin x}{\cos x(1 + \sin x)} = \frac{1}{\cos x} = \sec x
   $$
Thus, the statement is **true**.

### d) Prove that 
$$
\frac{\tan x \sin x}{\tan x + \sin x} = \frac{\tan x - \sin x}{\operatorname{can} x \sin x}
$$

**Solution:**
The expression contains a syntax error with "can". Assuming it should be "cos", let's rewrite it:
$$
\frac{\tan x \sin x}{\tan x + \sin x} = \frac{\tan x - \sin x}{\cos x \sin x}
$$
1. Start with the left-hand side:
   $$
   = \frac{\frac{\sin x}{\cos x} \sin x}{\frac{\sin x}{\cos x} + \sin x} = \frac{\frac{\sin^2 x}{\cos x}}{\frac{\sin x + \sin x \cos x}{\cos x}} = \frac{\sin^2 x}{\sin x + \sin x \cos x}
   $$
2. Simplify:
   $$
   = \frac{\sin^2 x}{\sin x(1 + \cos x)} = \frac{\sin x}{1 + \cos x}
   $$
3. Now for the right-hand side:
   $$
   = \frac{\frac{\sin x}{\cos x} - \sin x}{\cos x \sin x} = \frac{\frac{\sin x - \sin x \cos x}{\cos x}}{\cos x \sin x} = \frac{\sin x(1 - \cos x)}{\cos^2 x \sin x} = \frac{1 - \cos x}{\cos^2 x}
   $$
4. The two sides are not equal, thus the statement is **false**.

### e) Prove that 
$$
\frac{\tan x + \sec x}{\sec x\left(1 + \frac{\tan x}{\sin x}\right)} = 1
$$

**Solution:**
1. Start with the left-hand side:
   $$
   = \frac{\tan x + \sec x}{\sec x\left(1 + \frac{\tan x}{\sin x}\right)}
   $$
2. Rewrite $$\tan x$$ and $$\sec x$$:
   $$
   = \frac{\frac{\sin x}{\cos x} + \frac{1}{\cos x}}{\sec x\left(1 + \frac{\frac{\sin x}{\cos x}}{\sin x}\right)} = \frac{\frac{\sin x + 1}{\cos x}}{\sec x\left(1 + \frac{1}{\cos x}\right)}
   $$
3. Simplify:
   $$
   = \frac{\sin x + 1}{\cos x \cdot \sec x(1 + \frac{1}{\cos x})} = \frac{\sin x + 1}{(1)(1 + \frac{1}{\cos x})} = \frac{\sin x + 1}{1 + \frac{1}{\cos x}}
   $$
4. The two sides are not equal, thus the statement is **false**.

### Summary of Results:
- a) True
- b) True
- c) True
- d
a) True b) True c) True d) False e) False

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