To simplify, recognize that $\displaystyle 1-\cos^2\theta$ is a reworking on $\displaystyle \sin^2+\cos^2=1$, meaning that $\displaystyle 1-\cos^2\theta=\sin^2\theta$.

Plug that into our given equation:

$\displaystyle \frac{1}{1-\cos^2\theta}=\frac{1}{\sin^2\theta}$

Remember that $\displaystyle \csc\theta=\frac{1}{\sin\theta}$, so $\displaystyle \frac{1}{\sin^2\theta}=\csc^2\theta$.

Recognize that $\displaystyle 1-\cos^2\theta$ is a reworking on $\displaystyle \sin^2+\cos^2=1$, meaning that $\displaystyle 1-\cos^2\theta=\sin^2\theta$.

Plug that in to our given equation:

$\displaystyle \frac{1-\cos^2\theta}{\sin\theta\cos\theta}=\frac{\sin^2\theta}{\sin\theta\cos\theta}$

Notice that one of the $\displaystyle \sin\theta$'s cancel out.

$\displaystyle \frac{\sin\theta}{\cos\theta}=\tan\theta$.

$\displaystyle \frac{1}{(cos (33^{\circ}))^{2}} - (tan (33^{\circ}))^{2} = (sec (33^{\circ}))^{2}- (tan (33^{\circ}))^{2}$

Recall the Pythagorean Identity:

$\displaystyle 1 + (tan x)^{2} = (sec x)^{2}$

We can rearrange the terms:

$\displaystyle 1 = (sec x)^{2} - (tan x)^{2}$

This is exactly what our original equation looks like, so the answer is 1.

There are a couple valid strategies for solving this problem. The simplest is to first factor out $\displaystyle \cos x\sin x$ from both sides. This leaves us with:

$\displaystyle \cot^{2}x\cos x\sin x + \cos x \sin x = (\cos x \sin x)(cot^{2}x + 1)$

Next, substitute with the known identity $\displaystyle \cot^{2}x + 1=\csc^{2}x$ to get:

$\displaystyle \cos x\sin x (\csc^{2}x)$

From here, we can eliminate the quadratic by converting:

$\displaystyle \csc^{2}x= \frac{1}{\sin^{2}x}$

giving us

$\displaystyle \frac{\cos x\sin x}{\sin^{2}x}=\frac{\cos x}{\sin x}= \cot x$

Thus,

$\displaystyle \cot^{2}x\cos x\sin x+\cos x\sin x = \cot x$

The expression $\displaystyle (1-\cos a)(1+\cos a)$ represents a difference of squares. In this case, the product is $\displaystyle 1-\cos^{2}a$ (remember that 1 is also a perfect square).

One Pythagoran identity for trigonometric functions is: 

$\displaystyle 1-\cos^{2}a = \sin^{2}a$

Thus, we can say that the most simplified version of the expression is $\displaystyle \sin^{2}a$.

Write the Pythagorean Identity.

$\displaystyle sin^2(\theta)+cos^2(\theta)=1$

Substitute the value of $\displaystyle sin(\theta)$ and solve for $\displaystyle cos(\theta)$.

$\displaystyle (\frac{3}{5})^2+cos^2(\theta)=1$

$\displaystyle cos^2(\theta)=1-(\frac{3}{5})^2$

$\displaystyle cos^2(\theta)=1-(\frac{3}{5})^2= \frac{25}{25}-\frac{9}{25}=\frac{16}{25}$

$\displaystyle cos(\theta)=\pm \sqrt{\frac{16}{25}}=\pm \frac{4}{5}$

Since the cosine is in the second quadrant, the correct answer is:

$\displaystyle cos(\theta)=-\frac{4}{5}$

According to the Pythagorean identity

$\displaystyle sin^2x+cos^2x=1$,

the right hand side of this equation can be rewritten as $\displaystyle sin^2x$. This yields the equation

$\displaystyle sinxcosxtanx =sin^2x$ .

Dividing both sides by $\displaystyle sinx$ yields:

$\displaystyle cosxtanx=sinx$ .

Dividing both sides by $\displaystyle cosx$ yields:

$\displaystyle tanx=\frac{sinx}{cosx}$ .

This is precisely the definition of the tangent function; since the domain of $\displaystyle tanx$ consists of all real numbers, the values of $\displaystyle x$ which satisfy the original equation also consist of all real numbers. Hence, the correct answer is 

$\displaystyle \mathbb{R}=(-\infty, \infty )$.

By the Pythagorean identity, the first two terms simplify to 1:

$\displaystyle \sin^2x+\cos^2x+\tan^2x=1+\tan^2x$.

Dividing the Pythagorean identity by $\displaystyle \cos^2x$ allows us to simplify the right-hand side.

$\displaystyle \frac{\sin^2x}{\cos^2x}+\frac{\cos^2x}{\cos^2x}=\frac{1}{\cos^2x}$

$\displaystyle \tan^2x+1=\sec^2x$

Step 1: Recall the trigonometric identity that has sine and cosine in it...

$\displaystyle sin^2x+cos^2x=1$

The sum is equal to 1.

Using the Pythagorean Identity

$\displaystyle \sin^2+\cos^2=1$,

one can solve for $\displaystyle \cos^2(\theta)$ by plugging in $\displaystyle (1/4)^2$ for $\displaystyle \sin^2$.

Solving for $\displaystyle \cos^2(\theta)$, you get it equal to $\displaystyle 15/16$.

Taking the square root of both sides will get the correct answer of 

$\displaystyle \cos(\theta)=\pm\sqrt{\frac{15}{4}}$.