Question

1. $(1 - \text{cos}^2 x)(1 + \text{cot}^2 x) = 1$
2. $\frac{(\text{sin } x + \text{cos } x)^2}{\text{sin } x \text{ cos } x} = 2 + \text{sec } x \text{ csc } x$
3. $\frac{\text{sin } x}{\text{sin } x + \text{cos } x} = \frac{\text{tan } x}{1 + \text{tan } x}$
4. $\text{sec } x = \text{sin } x (\text{cot } x + \text{tan } x)$
5. $\tan x sec^2 x - \tan^3 x = \frac{sec x}{csc x}$

Explanation:

Let's solve problem 6: \((1 - \cos^2 x)(1 + \cot^2 x) = 1\)

Step 1: Use the Pythagorean identity for sine and cosine

We know that \(\sin^2 x + \cos^2 x = 1\), so \(1 - \cos^2 x=\sin^2 x\).

Step 2: Use the Pythagorean identity for cotangent and cosecant

We know that \(1 + \cot^2 x=\csc^2 x\), and \(\csc x=\frac{1}{\sin x}\), so \(\csc^2 x = \frac{1}{\sin^2 x}\).

Step 3: Substitute the identities into the left - hand side of the equation

Substitute \(1 - \cos^2 x=\sin^2 x\) and \(1 + \cot^2 x=\csc^2 x=\frac{1}{\sin^2 x}\) into \((1 - \cos^2 x)(1 + \cot^2 x)\):
\((1 - \cos^2 x)(1 + \cot^2 x)=\sin^2 x\times\frac{1}{\sin^2 x}\)

Step 4: Simplify the expression

\(\sin^2 x\times\frac{1}{\sin^2 x} = 1\)

Answer:

The left - hand side \((1 - \cos^2 x)(1 + \cot^2 x)\) simplifies to \(1\), which is equal to the right - hand side. So the identity \((1 - \cos^2 x)(1 + \cot^2 x)=1\) is proven.