I have a couple questions that I need a little direction on.

1:

Prove that $\displaystyle \sin^2 x(1+\cot^2 x)=1$

Method:

I figured out that I should convert $\displaystyle \sin^2 x(1+\cot^2 x)=1$ to $\displaystyle \sin^2 x (\csc^2 x)=1$. But beyond that, I'm a little lost....I know there is more to be done though. I just don't quite understand how reciprocal and Pythagorean identities are related, I know that $\displaystyle \sin x \csc x=1$ but I'm not sure if the same holds true for their Pythagorean counterparts or if that would be a valid proof.

2:

Prove that $\displaystyle \tan x(\cot x +\tan x)=\sec^2 x$.

Method:

I made the equation $\displaystyle \tan x(\cot x +\tan x)=\sec^2 x$ into $\displaystyle \tan x(\frac{1}{\tan x} + \tan x)= sec^2 x$.

Does the equation become $\displaystyle 1+\tan^2=\sec^2$, or am I missing something?

I think I have the right answer, as I see that the Pythagorean identity for $\displaystyle \sec^2 x$ coinsides with my converted equation.

3:

Find the exact value of $\displaystyle \cos 105$ with out using tables or a calculator.

Be sure to show your work.

Method:

Since the question asks for the value of a unusual degree measure I have to use an addition formula, namely: $\displaystyle \cos(\alpha - \beta)=\cos\alpha \cos \beta+\sin \alpha \sin \beta$.

For \alpha I used value 135 and for \beta I used 30 so $\displaystyle \cos(\alpha - \beta) \rightarrow \cos(135-30)$, therefore the whole formula changes to become: $\displaystyle \cos(135 - 30)=\cos 135 \cos 30 + \sin 135 \sin 30$.

So, $\displaystyle \cos 135 \cos 30 + \sin 135 \sin 30 \rightarrow -\frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \frac{1}{2} \rightarrow \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} \rightarrow \frac{\sqrt{2} +\sqrt{6}}{4}$, have I gotten the right answer?

4:

Find the direct value of $\displaystyle \tan 15$ without using tables or a calculator.

Be sure to to show your work.

Method:

The book said nothing about the addition formula for $\displaystyle \tan$ so I searched the internet and found this:$\displaystyle \tan(\alpha-\beta)= \frac{\tan \alpha - \tan \beta}{1+ \tan\alpha \tan\beta}$

I decided to use this formula since I could figure out no other to use, I entered $\displaystyle \tan 45$ for $\displaystyle \alpha$ and $\displaystyle \tan 30$ for $\displaystyle \beta$, but I dont know where to go from here....

Thanks.