Maths help please? (1 Viewer)

LoveHateSchool · May 16, 2011

I have trouble with a few trig questions..

1) cosecθsecθ-cotθ
2) 1-cos2θ over 1-sin2 θ
3) Sin2x+cos2x over tan2x
4) (1+tan2 θ)(1-sin2 θ)
5) 1-cos2 θ over cosθ

I would love any help on any of them! I know the identities but have trouble knowing which to use and such :S

maths lover · May 16, 2011

what do u need just simplifying

LoveHateSchool · May 16, 2011

Yes, just to simplify them

slyhunter · May 16, 2011

1.

$\frac{1}{sin\theta } \cdot \frac{1}{cos \theta } - \frac{cos \theta }{sin \theta }$

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

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

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

$=\frac{sin \theta }{cos \theta }$

$=tan \theta$

Is q2 $cos 2\theta$ and $sin^{2}\theta$ ?

AAEldar · May 16, 2011

slyhunter · May 16, 2011

^herp derp

LoveHateSchool · May 16, 2011

Thanks for the help on the first one sly, and sorry the second one is 1-cos squared theta over 1-sin squared theta.

LoveHateSchool · May 16, 2011

In fact all the 2's I wrote are meant to be a little two as in squared *blushes*

SpiralFlex · May 16, 2011

1. $\frac{1}{\sin\theta}*\frac{1}{\cos\theta} - \frac{\cos\theta}{\sin\theta}$

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

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

$= \frac{ \sin \theta}{\cos \theta}$

$= \tan \theta$

2. $\frac{\sin^2 x}{\cos^2 x}$

$=\tan^2 \theta$

3. $\frac{1}{\tan^2 \theta}$

$=cot^2 \theta$

4. $sec^2 x * \cos^2 x$

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

$= 1$

5. $\frac{\sin^2 \theta}{\cos\theta}$

If you are not familiar with your identities at least remember the first identity. $\sin^2 \theta + \cos^2 \theta = 1$
[Note]: You should put ^ to avoid future confusion of $\sin^2 x$ or $\sin 2x$

From $\sin^2 \theta + \cos^2 \theta = 1$ . You can derive the rest of your identities.

slyhunter · May 16, 2011

2.

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

$=tan ^{2}\theta$

Herp derp

slyhunter · May 16, 2011

3.

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

$= cot ^{2}x$

LoveHateSchool · May 16, 2011

^Thank you all so much! Unfortunately I must spread rep, with the second last one, I've rewritten it as (1+ tan squared teta) (cos sqaured theta), what do I do from there?

slyhunter · May 16, 2011

Spi did it just then, answer is 1

SpiralFlex · May 16, 2011

LoveHateSchool said:
^Thank you all so much! Unfortunately I must spread rep, with the second last one, I've rewritten it as (1+ tan squared teta) (cos sqaured theta), what do I do from there?

-Points to my post-

LoveHateSchool · May 16, 2011

Oh I see Spi's now, I love you all and I'll make sure to write ^2 in future, I didn't release that copying it out ofw rod got rid of superscript whatever.

slyhunter · May 16, 2011

Use LaTeX

LoveHateSchool · May 16, 2011

^I will have to

SpiralFlex · May 16, 2011

LoveHateSchool said:
^I will have to

Remember to use it.

AAEldar · May 17, 2011

Lol, I thought there was something wrong when I "simplified" 2 and it came out just as ugly.

AAEldar · May 17, 2011

SpiralFlex said:
5.
$\frac{1-cos^2\theta}{cos\theta} \\ =\frac{\sin^2 \theta}{\cos\theta} \\ =sin\theta tan\theta$

I'd simplify it to that.

Maths help please? (1 Viewer)

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