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. \( \tan ^{2} \theta+1=\frac{1}{\cos ^{2} \theta} \)

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, \[ \tan^2\theta+1=\frac{1}{\cos^2\theta}. \]