1. ## Trig equation help please!

I think I'm having a massive blonde moment...

How would I begin?

edit: also this question:

Here's my working so far:

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

$\frac{1+\frac{sin^2x}{1-sin^2x}}{1-sin^2x}$

I'm stuck here. I don't actually know what to do with multi-layered fractions, never been taught it.

So clearly, I have to multiply by $1-sin^2x$

How do I get there?

Also, this question:

part b,

Here's my working, I'm starting to think the question is wrong..?

$\frac{1}{cos\theta} + \frac{sin\theta}{cos\theta}=\frac{cos\theta}{1-sin\theta}$

$\frac{cos\theta+sin\theta cos\theta}{cos^2\theta}$

$\frac{cos\theta(1+sin\theta)}{1-sin^2\theta}$

What do I do from here?

2. ## Re: Trig equation help please!

9)

let $u=x^2$

$\sin(u)=\dfrac 1 2$ for $-\pi \leq u \leq \pi$

can you solve that? There are two solutions in between $-\pi$ and $\pi$. You should be able to solve that.
Then solve $u=x^2$ This will give you 4 solutions total.

10)

hint: $1+\tan^2(x)=\sec^2(x)$

2nd hint: (you shouldn't need this one) $\cos^4(x)=\left(\cos^2(x)\right)^2=\left(1-sin^2(x)\right)^2$

11)

$\dfrac 1 {\cos(\theta)}+\tan(\theta)=\dfrac{1+\sin(\theta)} {cos(\theta)}$

Now multiply that by $1$ in the form of $\dfrac{1-\sin(\theta)}{1-\sin(\theta)}$ and simplify it.

3. ## Re: Trig equation help please!

Originally Posted by romsek
x

Can I just confirm something for question 10 though?

$\frac{\frac{1}{cos^2x}}{cos^2x}$

That simplifies to $\frac{1}{(cos^2x)^2}$ right? That's how I get to the answer? Sorry - I don't fully understand multiple layered fractions.

Edit: For 11, $\frac{cos\theta}{1-sin\theta} = \frac{1+sin\theta}{cos\theta}$? Confused :s

4. ## Re: Trig equation help please!

Originally Posted by Bude8

Can I just confirm something for question 10 though?

$\frac{\frac{1}{cos^2x}}{cos^2x}$

That simplifies to $\frac{1}{(cos^2x)^2}$ right? That's how I get to the answer? Sorry - I don't fully understand multiple layered fractions.
you need to get the basics down cold. Yes that's correct.

Edit: For 11, $\frac{cos\theta}{1-sin\theta} = \frac{1+sin\theta}{cos\theta}$? Confused :s
sigh.. it's just algebra

$\dfrac 1 {\cos(\theta)} + \tan(\theta) = \dfrac 1 {\cos(\theta)}+\dfrac{\sin(\theta)}{\cos(\theta)}=$

$\dfrac{1+\sin(\theta)}{\cos(\theta)}=\dfrac{1+ \sin(\theta)}{\cos(\theta)}\cdot \dfrac{1- \sin(\theta)}{1- \sin(\theta)}=$

$\dfrac{1-\sin^2(\theta)}{\cos(\theta)(1-\sin(\theta))}=\dfrac{\cos^2(\theta)}{\cos(\theta) (1-\sin(\theta))}=$

$\dfrac{\cos(\theta)}{1-\sin(\theta)}$