EXERCISE 5Prove The Following Identities:1) $\sin X \cdot \cos X \cdot \tan X = 1 - \cos^2 X$2) $\cos^3 X + \cos X \cdot \sin^2 X = \cos X$3) $(\sin X + \cos X)^2 + (\sin X - \cos X)^2 = 2$4) $\left(1 - \sin^2

In this article, we will prove four trigonometric identities using various mathematical techniques. These identities are essential in trigonometry and are used to simplify complex expressions involving sine, cosine, and tangent functions.

Identity 1: $\sin x \cdot \cos x \cdot \tan x = 1 - \cos^2 x$

To prove this identity, we can start by expressing $\tan x$ in terms of $\sin x$ and $\cos x$. We know that $\tan x = \frac{\sin x}{\cos x}$. Substituting this expression into the left-hand side of the identity, we get:

$\sin x \cdot \cos x \cdot \tan x = \sin x \cdot \cos x \cdot \frac{\sin x}{\cos x}$

Simplifying this expression, we get:

$\sin x \cdot \cos x \cdot \tan x = \sin^2 x$

Now, we can use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ to rewrite $\sin^2 x$ as $1 - \cos^2 x$. Substituting this expression into the previous equation, we get:

$\sin x \cdot \cos x \cdot \tan x = 1 - \cos^2 x$

Therefore, we have proved the first identity.

Identity 2: $\cos^3 x + \cos x \cdot \sin^2 x = \cos x$

To prove this identity, we can start by factoring out $\cos x$ from the left-hand side of the equation:

$\cos^3 x + \cos x \cdot \sin^2 x = \cos x (\cos^2 x + \sin^2 x)$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x + \sin^2 x$ as $1$. Substituting this expression into the previous equation, we get:

$\cos^3 x + \cos x \cdot \sin^2 x = \cos x (1)$

Simplifying this expression, we get:

$\cos^3 x + \cos x \cdot \sin^2 x = \cos x$

Therefore, we have proved the second identity.

Identity 3: $(\sin x + \cos x)^2 + (\sin x - \cos x)^2 = 2$

To prove this identity, we can start by expanding the squares on the left-hand side of the equation:

$(\sin x + \cos x)^2 + (\sin x - \cos x)^2 = \sin^2 x + 2 \sin x \cos x + \cos^2 x + \sin^2 x - 2 \sin x \cos x + \cos^2 x$

Simplifying this expression, we get:

$(\sin x + \cos x)^2 + (\sin x - \cos x)^2 = 2 \sin^2 x + 2 \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $2 \sin^2 x + 2 \cos^2 x$ as $2$. Substituting this expression into the previous equation, we get:

$(\sin x + \cos x)^2 + (\sin x - \cos x)^2 = 2$

Therefore, we have proved the third identity.

Identity 4: $\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1$

To prove this identity, we can start by expanding the square on the left-hand side of the equation:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - 2 \sin^2 x + \sin^4 x + \sin^2 x \cdot \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - 2 \sin^2 x + \sin^4 x + \sin^2 x \cdot \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\sin^2 x \cdot \cos^2 x$ as $\sin^2 x (1 - \sin^2 x)$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - 2 \sin^2 x + \sin^4 x + \sin^2 x (1 - \sin^2 x)$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - 2 \sin^2 x + \sin^4 x + \sin^2 x - \sin^4 x$

Combining like terms, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

In this article, we will prove four trigonometric identities using various mathematical techniques. These identities are essential in trigonometry and are used to simplify complex expressions involving sine, cosine, and tangent functions.

Identity 4: $\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1$

To prove this identity, we can start by expanding the square on the left-hand side of the equation:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - 2 \sin^2 x + \sin^4 x + \sin^2 x \cdot \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - 2 \sin^2 x + \sin^4 x + \sin^2 x \cdot \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\sin^2 x \cdot \cos^2 x$ as $\sin^2 x (1 - \sin^2 x)$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - 2 \sin^2 x + \sin^4 x + \sin^2 x (1 - \sin^2 x)$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - 2 \sin^2 x + \sin^4 x + \sin^2 x - \sin^4 x$

Combining like terms, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $1 - \sin^2 x$ as $\cos^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = \cos^2 x$

Using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, we can rewrite $\cos^2 x$ as $1 - \sin^2 x$. Substituting this expression into the previous equation, we get:

$\left(1 - \sin^2 x\right)^2 + \sin^2 x \cdot \cos^2 x = 1 - \sin^2 x$

Simplifying this expression, we get:

$\left(1 - \sin^