Simplifying calculus homework problems

[hmhmherath]

Can anyone simplify any of these questions?

 

[Greg]

Hello hmhmherath and welcome to MHB! :D

As per forum rules, in future please post at most two problems per thread.

Let's take these one at a time. For the first one (problem 11), do you know the formula for the sum of a geometric series?

 

[Prove It]

Can anyone simplify any of these questions?

$\displaystyle \begin{align*} \frac{1}{\cos^2{(x)}} &= 2\,\sqrt{3}\tan{(x)}- 2 \\ 1 &= 2\,\sqrt{3} \sin{(x)}\cos{(x)} - 2\cos^2{(x)} \\ 1 &= \sqrt{3}\sin{(2\,x)} - 1 - \cos{(2\,x)} \\ 2 + 2\cos{(2\,x)} &= \sqrt{3}\sin{(2\,x)} \\ 4 + 8\cos{(2\,x)} + 4\cos^2{(2\,x)} &= 3 \sin^2{(2\,x)} \\ 4 + 8\cos{(2\,x)} + 4\cos^2{(2\,x)} &= 3 - 3\cos^2{(2\,x)} \\ 7\cos^2{(2\,x)} + 8\cos{(2\,x)} + 1 &= 0 \end{align*}$

Now solve the resulting quadratic.

 

[Greg]

$$\frac{1}{\cos^2\theta}=2\sqrt3\tan\theta-2$$

$$\sec^2\theta=2\sqrt3\tan\theta-2$$

$$\tan^2\theta+1=2\sqrt3\tan\theta-2$$

$$\tan^2\theta-2\sqrt3\tan\theta+3=0$$

Can you solve that quadratic for $\tan\theta$ ?