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 

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