Simplif. $$ an ^{2} x\left(1-\sin ^{2} x ight) $$ 2. $$ (3-3 \sin heta)(3+3 \sin heta) $$ 3. $$ \left(\frac{1}{ an x}+ an x ight) \cdot(\sin x \cdot \cos x) $$ Prii! 4. $$ \cos ^{3} x+\cos x \cdot \sin ^{2} x=\cos x $$ 5. $$ an ^{2} heta+1=\frac{1}{\cos ^{2} heta} $$

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Simplif. $$ 	an ^{2} x\left(1-\sin ^{2} xight) $$ 2. $$ (3-3 \sin 	heta)(3+3 \sin 	heta) $$ 3. $$ \left(\frac{1}{	an x}+	an xight) \cdot(\sin x \cdot \cos x) $$ Prii! 4. $$ \cos ^{3} x+\cos x \cdot \sin ^{2} x=\cos x $$ 5. $$ 	an ^{2} 	heta+1=\frac{1}{\cos ^{2} 	heta} $$

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1. We start with
$$
\tan^2 x \left( 1-\sin^2 x \right).
$$
Recall that
$$
1-\sin^2 x=\cos^2 x.
$$
Thus the expression becomes
$$
\tan^2 x \cdot \cos^2 x.
$$
Since
$$
\tan^2 x=\frac{\sin^2 x}{\cos^2 x},
$$
we substitute to obtain
$$
\frac{\sin^2 x}{\cos^2 x}\cdot\cos^2 x=\sin^2 x.
$$

2. Consider
$$
(3-3\sin \theta)(3+3\sin \theta).
$$
Factor out $$3$$ from both parentheses:
$$
=3(1-\sin \theta)\cdot 3(1+\sin \theta)=9(1-\sin \theta)(1+\sin \theta).
$$
Using the difference of squares, we know that
$$
(1-\sin \theta)(1+\sin \theta)=1-\sin^2 \theta.
$$
But
$$
1-\sin^2 \theta=\cos^2 \theta.
$$
Thus, the simplified expression is
$$
9\cos^2 \theta.
$$

3. Evaluate
$$
\left(\frac{1}{\tan x}+\tan x\right) \cdot (\sin x\cdot \cos x).
$$
Express $$\tan x$$ in terms of sine and cosine:
$$
\tan x=\frac{\sin x}{\cos x}\quad\text{and}\quad \frac{1}{\tan x}=\frac{\cos x}{\sin x}.
$$
Substitute these into the expression:
$$
\left(\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}\right) \cdot (\sin x\cos x).
$$
Distribute the factor $$\sin x\cos x$$ term by term:
$$
\frac{\cos x}{\sin x}\cdot \sin x\cos x = \cos^2 x,
$$
$$
\frac{\sin x}{\cos x}\cdot \sin x\cos x = \sin^2 x.
$$
Adding these gives
$$
\cos^2 x+\sin^2 x=1.
$$

4. Simplify
$$
\cos^3 x + \cos x\cdot\sin^2 x.
$$
Factor out $$\cos x$$:
$$
= \cos x \left( \cos^2 x + \sin^2 x \right).
$$
Since
$$
\cos^2 x+\sin^2 x=1,
$$
the expression becomes
$$
\cos x.
$$

5. Show that
$$
\tan^2\theta+1=\frac{1}{\cos^2\theta}.
$$
Recall the trigonometric identity
$$
\tan^2\theta+1=\sec^2\theta.
$$
And since
$$
\sec\theta=\frac{1}{\cos\theta},
$$
it follows that
$$
\sec^2\theta=\frac{1}{\cos^2\theta}.
$$
Thus,
$$
\tan^2\theta+1=\frac{1}{\cos^2\theta}.
$$
1. $$ \tan^2 x (1 - \sin^2 x) = \sin^2 x $$ 2. $$ (3 - 3\sin \theta)(3 + 3\sin \theta) = 9\cos^2 \theta $$ 3. $$ \left(\frac{1}{\tan x} + \tan x\right) \cdot (\sin x \cdot \cos x) = 1 $$ 4. $$ \cos^3 x + \cos x \cdot \sin^2 x = \cos x $$ 5. $$ \tan^2 \theta + 1 = \frac{1}{\cos^2 \theta} $$

Simplif. \( \tan ^{2} x\left(1-\sin ^{2} x\right) \) 2. \( (3-3 \sin \theta)(3+3 \sin \theta) \) 3. \( \left(\frac{1}{\tan x}+\tan x\right) \cdot(\sin x \cdot \cos x) \) Prii! 4. \( \cos ^{3} x+\cos x \cdot \sin ^{2} x=\cos x \) 5. \( \tan ^{2} \theta+1=\frac{1}{\cos ^{2} \theta} \)

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