$\tan\theta=\sec\theta$? Test questions 
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*If $\theta$ is in quadrant $\text{I}$ and $\tan(\theta) = 0.6$ then $\sec(\theta) = $?
This seems pretty easy to me: $$\tan^2(\theta)-\sec^2(\theta)=1
\\ -\sec^2(\theta)=.64
\\ \sec(\theta)=8$$ 

*Another one, $\cos(\theta)=\sin(2\theta)$
Should that be $\cos^2(\theta)+\sin^2(\theta)=1/2$? Which would be 0.5 

*For all angles $\theta$, $\cos(-\theta)$ = $\cos(\theta)$
$\ \ \ \ \ \ \ $I said false how can a negative angle be equal to a positive one?


*If $\sin^2(\theta)= 0.5$ then $\sin^2(\theta) = \cos^2(\theta)$
I said false because $\sin$ and $\cos$ can never equal 0.5 together but it was wrong.
This answer is wrong but I don't understand why. 


I am seriously considering changing majors, I know most people here will think I am an idiot, lazy or whatever else for taking such simple high school level math courses in college but I really am having trouble with it. I keep making mistakes no matter what I do I can never get better than a D or C on a test. So I am trying to evaluate the mistakes I made on the test so I don't make them again, but inevitably I will.
 A: 
(2) If the problem asked you for the angle $\theta$ at which $\cos \theta = \sin(2\theta)$, 

then think about what you know about a 30, 60, 90 right triangle: $$\cos(30^\circ) = \sin(2\cdot 30^\circ) = \sin(60^\circ) = \frac {\sqrt{3}}{2}$$
So theta would be $30^\circ$. 

(4) If $\sin^2 \theta = .5$ then $\sin^2 \theta = \cos^2 \theta$.

This is true, since if $\sin^2 \theta = .5 = \frac 12$, then we know that 
$$ \sin \theta = \pm\sqrt {\frac 12} = \pm \frac {\sqrt 2}{2} = \pm \frac {1}{\sqrt 2} = \pm \cos \theta$$  (That means the reference angle is $45^\circ$.)Hence $\cos^2 \theta = \sin^2 \theta = 0.5$.
A: You need to read the questions more carefully.  Checking the typing would make them easier to read, as there are many typos.  Setting them in $\LaTeX$ (see the FAQ) would help a lot, too.
For the first, the identity is $1+tan^2\theta=sec^2\theta$.  Your first equality is incorrect as $\tan^2\theta \neq 1$.  The dash after $1$ looks like a minus sign, but it seems to be a separator.  How about line feeds?  Then you lost a decimal point at the end.
For the second, what is sin2theta?  $\sin^2 \theta$ or $\sin 2\theta$?
Please review the question and make it legible.
A: Draw a triangle in the first quadrant with $\tan(\theta) = 3/5$.  It will have height 3, base 5 and hypoteneuse $\sqrt{34}$.   When I compute $\sec(\theta)$
I get $\sqrt{34}/5$.  Drawing the appropriate picture makes this worlds simpler.  
A: part 1:
Identity is
$$\sec^2\theta -\tan^2\theta =1$$
$$\sec^2\theta -0.36 =1$$
$$\sec\theta =\sqrt{1.36}\implies\sec\theta=1.1662$$
part 2:
$$\cos \theta=\sin2\theta\implies\cos\theta=\cos\left(\dfrac\pi2-2\theta\right)\implies\theta=\dfrac\pi2-2\theta\implies\theta=\dfrac\pi6$$
$$\cos^2(\dfrac \pi6)+\sin^2(\dfrac \pi6)=1$$
part3:
$$\cos(-\theta)\implies\cos(0^\circ-\theta)\implies\cos 0^\circ\cdot\cos\theta+\sin 0^\circ\cdot\sin\theta\implies\cos\theta$$
so: $$\cos(-\theta)=\cos\theta$$
part4:
if $\sin^2\theta=\dfrac12\implies\sin^2\theta=\left(\dfrac{1}{\sqrt{2}}\right)^2\implies\sin^2\theta=\sin^2\left(\dfrac{\pi}{4}\right)$
if
$\sin^2\theta=\sin^2\alpha$  then general solution of $\theta=n\pi\pm\alpha\;\;,n \in Z$.
so:
$$\theta=n\pi\pm\dfrac{\pi}{4}\;\;,n \in Z$$
so there are not $\sin\theta=\cos\theta=0.5\;\;while\;\;\sin^2\theta=\cos^2\theta=0.5$
